The transformations on the nome and Landen's transformation

Question

Could someone please explain how to transform the nome $q = e^{-\pi K'/K}$ from $q^2$ to $q$ and then to $-q$? In other words, how does changing $q^2$ to $q$ and then $q$ to $-q$ affect $k$ and $K$. Here, $k$ is the elliptic modulus, $K = K(k)$ is the complete elliptic integral of the first kind, and $K' = K(k')$, where $k' = \sqrt{1 - k^2}$ is the complementary modulus. I know that the transformation of $q$ is somehow connected to Landen's transformation, but I do not understand how these two concepts are related. Thank you in advance.

Answer 1

I tend to look at this using Jacobi thetanull functions, as these are usually expressed in terms of the nome $q$: $$\begin{align} \vartheta_2(q) &= q^{1/4}\sum_{n\in\mathbb{Z}} q^{n(n+1)} & \vartheta_3(q) &= \sum_{n\in\mathbb{Z}} q^{n^2} & \vartheta_4(q) &= \sum_{n\in\mathbb{Z}} (-q)^{n^2} \end{align}$$

As to the connection of your question with Landen's transformation, let me just remark that Landen's transformation gives rise to the arithmetic-geometric mean iteration step, which in turn resembles the transformation of $\vartheta_3^2$ and $\vartheta_4^2$ under $q\mapsto q^2$. To answer your question, it suffices to do that step backwards. I will focus on the Jacobi thetanulls here, and thus get along without considering integrals and Landen's transformation as such. Expect some ambiguity from inverse powers.

The formulae we need are $$\begin{align} \vartheta_2(-q) &= \pm\sqrt{\pm\mathrm{i}}\,\vartheta_2(q) & \vartheta_2^2(q) &= 2\,\vartheta_3(q^2)\,\vartheta_2(q^2) \\ \vartheta_3(-q) &= \vartheta_4(q) & \vartheta_3^2(q) &= \vartheta_3^2(q^2) + \vartheta_2^2(q^2) \\ \vartheta_4(-q) &= \vartheta_3(q) & \vartheta_4^2(q) &= \vartheta_3^2(q^2) - \vartheta_2^2(q^2) \end{align}$$ The formulae for $q\mapsto-q$ are obvious, and the last one of those for $q^2\mapsto q$ has been proven elsewhere on this site. The remaining ones follow in a similar manner.

For the relations of the Jacobi thetanulls to the complete elliptic integral of the first kind, one only needs to keep in mind that $$\begin{align} k(q) &= \frac{\vartheta_2^2(q)}{\vartheta_3^2(q)} & k'(q) &= \frac{\vartheta_4^2(q)}{\vartheta_3^2(q)} & K\left(k(q)\right) &= \frac{\pi}{2}\vartheta_3^2(q) \end{align}$$

From those identities we can deduce $$\begin{align} k(-q) &= \pm\mathrm{i}\frac{k(q)}{k'(q)} & k'(-q) &= \frac{1}{k'(q)} \\ k(q) &= \pm\frac{2\sqrt{k(q^2)}}{1+k(q^2)} & k'(q) &= \frac{1-k(q^2)}{1+k(q^2)} \\\therefore\quad k(-q) &= \pm\mathrm{i}\frac{2\sqrt{k(q^2)}}{1-k(q^2)} & k'(-q) &= \frac{1+k(q^2)}{1-k(q^2)} \end{align}$$ and for the complete elliptic integral: $$\begin{align} K\left(k(-q)\right) &= k'(q)\,K\left(k(q)\right) \\ K\left(k(q)\right) &= \left(1+k(q^2)\right) K\left(k(q^2)\right) \\\therefore\quad K\left(k(-q)\right) &= \left(1-k(q^2)\right) K\left(k(q^2)\right) \end{align}$$ which is what you have asked for.