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Reading some Conformal Field Theory, I came across the following equation about the Jacobi Theta functions without any justification:

Let $$\theta_{2}(q)=\sum_{n \in \mathbb{Z}}q^{(n+\frac{1}{2})^{2}}$$ and $$\theta_{3}(q)=\sum_{n \in \mathbb{Z}}q^{n^{2}}$$ then:

$$\frac{\theta_{3}(q^{2})\theta_{3}(\overline{q}^{2})+\theta_{2}(q^{2})\theta_{2}(\overline{q}^{2})}{\theta_{3}(q)\theta_{3}(\overline{q})}=\frac{1}{\sqrt{2}}\sqrt{\left|\frac{\theta_{2}^{4}(q)}{\theta_{3}^{4}(q)}\right|+1+\left|1-\frac{\theta_{2}^{4}(q)}{\theta_{3}^{4}(q)}\right|}.$$

Indeed, this seems to hold true when I plot the two sides. Does somebody has a clue why this equation holds ? Any comments or any reference ?

Thanks !

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  • $\begingroup$ Does $\bar q$ denote $q^2$ or $e^{-\pi i\tau}$? $\endgroup$ Oct 10, 2022 at 16:51
  • $\begingroup$ This is the conjugate in this equation. $\endgroup$
    – Mathix
    Oct 10, 2022 at 18:10

1 Answer 1

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By using standard identities from the theory of Elliptic functions, you can prove it at least for the real nome $q\in(-1,1)$. There are actually many identities involving products of Theta functions. I will use these two $$ \theta_{3}^{2}(q^{2})+\theta_{2}^{2}(q^{2})=\theta_{3}^{2}(q) \quad\mbox{ and }\quad \theta_{2}^{4}(q)+\theta_{4}^{4}(q)=\theta_{3}^{4}(q). $$ Both identities are proven, for instance, in D. F. Lawden, Elliptic Functions and Applications, Applied Mathematical Sciences, vol. 80, Springer-Verlag, New York, 1989, as Equations 1.4.21 and 1.4.53.

For $q\in(-1,1)$, the identity from the question can be written as $$ 2\left(\theta_{3}^{2}(q^{2})+\theta_{2}^{2}(q^{2})\right)^{2}=\theta_{2}^{4}(q)+\theta_{3}^{4}(q)+|\theta_{3}^{4}(q)-\theta_{2}^{4}(q)|. $$ By the first identity mentioned above, $$ \mbox{LHS} = 2\theta_{3}^{4}(q), $$ while the second identity applied twice yields $$ \mbox{RHS} = \theta_{2}^{4}(q)+\theta_{3}^{4}(q)+\theta_{4}^{4}(q)=2\theta_{3}^{4}(q). $$

By inspection of the used identities, you may try to modify the approach for general $q\in\mathbb{C}$ with $|q|<1$.

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  • $\begingroup$ Thanks for the reference and the explanations ! $\endgroup$
    – Mathix
    Oct 12, 2022 at 13:22

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