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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.

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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 $\Bbb{C}^*$.

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  • $\begingroup$ I cannot yet follow this. In particular as the above f does not satisfy the diff equation on the nose it seems that g is not log(q)-periodic? $\endgroup$
    – user108998
    Apr 12, 2021 at 18:14
  • $\begingroup$ 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 $\log z$. $\endgroup$
    – reuns
    Apr 12, 2021 at 18:19
  • $\begingroup$ my apologies, i was confused! $\endgroup$
    – user108998
    Apr 12, 2021 at 18:23
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    $\begingroup$ 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 $0$. $\endgroup$
    – Alexandre Eremenko
    Apr 12, 2021 at 23:55
  • $\begingroup$ @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. $\endgroup$
    – reuns
    Apr 12, 2021 at 23:56

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