# Analytic function with q- difference equation involving theta

Consider the analytic space $$\mathbb{C}^{*}$$ with coordinate $$z$$. Let $$q$$ be some parameter with $$|q|<1$$ and define the analytic function $$\theta(z;q):=\sum_{n\in\mathbb{Z}}q^{\binom{n}{2}}(-z)^{n}.$$ Remark that this is I think usually denoted $$\theta_{11}$$, and is the theta function corresponding to the trivial two torsion point on the elliptic curve $$E_{q}=\mathbb{C}^{*}/q^{\mathbb{Z}}$$.

Question. Is there an analytic function on $$\mathbb{C}^{*}$$, denoted $$s(z)$$, with the property $$s(qz)-s(z)=\theta(z).$$ Moreover can it be expressed neatly as some $$q$$-series? What I would really like is a reference to a text where such identities are collected in bulk so I don´t have to spend too much time trying to invent them myself.

No, because the constant coefficient $$c_0=q^0=1$$ of your Laurent series is non-zero. Look at $$f(z) =c_0\frac{\log z}{\log q}+ \sum_{n\ne 0} \frac{q^{n(n-1)/2}}{q^n-1} (-z)^n, \qquad f(qz)-f(z)=\theta(z)\bmod \frac{c_0 2i\pi }{\log q}$$ Assuming your $$s(z)$$ exists let $$g(w)=s(e^w)-f(e^w)$$ which is entire, $$\log q$$ periodic, and $$g(w+2i\pi)= g(w)-c_0\frac{2i\pi}{\log q}$$ Therefore, $$\frac{g(w)-g(0)}{w}$$ is entire and bounded so that $$g(w)=g(0)+wg'(0)$$.
$$g(w+2i\pi)= g(w)-c_0\frac{2i\pi}{\log q}$$ gives that $$g'(0)\ne 0$$, contradicting that $$g$$ is $$\log q$$-periodic, contradicting that $$s(z)$$ is analytic on $$\Bbb{C}^*$$.
• What do you mean? $f(e^w)$ is entire and $f(e^{w+\log q} )-f(e^w) = \theta(e^w)$. The problem is that $f(e^w)$ is not $2i\pi$-periodic, there is a $2i\pi$ term coming from the continuation of $\log z$. Apr 12, 2021 at 18:19
• I did not understand why anything else except the first sentence was necessary. Since $s$ is assumed to be analytic in $C^*$ it has a Laurent expansion, and in $s(qz)-s(z)$, the constant term is evidently $0$. Apr 12, 2021 at 23:55
• @AlexandreEremenko True, and I missed that, I guess the point is that when the constant term is $0$ there is an obvious solution, and when it is not as you say there is an obvious contradiction. Apr 12, 2021 at 23:56