Elementary Approach to Modular Equations: Jacobi's Transformation Theory 4

Transformation of Elliptic Functions

The relation $$y = \frac{x}{M}\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{x^{2}}{\text{sn}^{2}\,4s\omega}}{1 - k^{2}x^{2}\text{sn}^{2}\,4s\omega}$$ and other variants of it $$\begin{align}1 - y &= (1 - x)\prod_{s = 1}^{(p - 1)/2}\dfrac{\left(1 - \dfrac{x}{\text{sn}(K - 4s\omega)}\right)^{2}}{1 - k^{2}x^{2}\,\text{sn}^{2}\,4s\omega}\notag\\ 1 + y &= (1 + x)\prod_{s = 1}^{(p - 1)/2}\dfrac{\left(1 + \dfrac{x}{\text{sn}(K - 4s\omega)}\right)^{2}}{1 - k^{2}x^{2}\,\text{sn}^{2}\,4s\omega}\notag\\ 1 - ly &= (1 - kx)\prod_{s = 1}^{(p - 1)/2}\frac{(1 - kx\,\text{sn}(K - 4s\omega))^{2}}{1 - k^{2}x^{2}\,\text{sn}^{2}\,4s\omega}\notag\\ 1 + ly &= (1 + kx)\prod_{s = 1}^{(p - 1)/2}\frac{(1 + kx\,\text{sn}(K - 4s\omega))^{2}}{1 - k^{2}x^{2}\,\text{sn}^{2}\,4s\omega}\notag\end{align}$$ as described in previous post lead to the differential equation $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ when $$M = (-1)^{(p - 1)/2}\prod_{s = 1}^{(p - 1)/2}\left(\frac{\text{sn}(K - 4s\omega)}{\text{sn}\,4s\omega}\right)^{2}$$ and $$l = k^{p}\prod_{s = 1}^{(p - 1)/2}\text{sn}^{4}(K - 4s\omega)$$
Putting $x = \text{sn}(u, k)$ in the differential equation we see that $y = \text{sn}(u/M, l)$. Hence the relation between $x$ and $y$ in effect expresses $\text{sn}(u/M, l)$ in terms of elliptic functions of a different but related modulus $k$.

Thus we obtain $$\text{sn}\left(\frac{u}{M}, l\right) = \frac{\text{sn}\,u}{M}\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}\,4s\omega}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,4s\omega}$$ Using the expressions of $1 - y, 1 + y, 1 - ly, 1 + ly$ it is easy to see that $$\begin{align}\sqrt{1 - y^{2}} &= \sqrt{1 - x^{2}}\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{x^{2}}{\text{sn}^{2}(K - 4s\omega)}}{1 - k^{2}x^{2}\,\text{sn}^{2}\,4s\omega}\notag\\ \sqrt{1 - l^{2}y^{2}} &= \sqrt{1 - k^{2}x^{2}}\prod_{s = 1}^{(p - 1)/2}\frac{1 - k^{2}x^{2}\,\text{sn}^{2}(K - 4s\omega)}{1 - k^{2}x^{2}\,\text{sn}^{2}\,4s\omega}\notag\end{align}$$ In the last relation if we put $x = 1$ so that $y = 1$ we get the expression for $l'$ as $$l' = k'\prod_{s = 1}^{(p - 1)/2}\frac{1 - k^{2}\,\text{sn}^{2}(K - 4s\omega)}{1 - k^{2}\,\text{sn}^{2}\,4s\omega} = k'\prod_{s = 1}^{(p - 1)/2}\frac{\text{dn}^{2}(K - 4s\omega)}{\text{dn}^{2}\,4s\omega} = k'\prod_{s = 1}^{(p - 1)/2}\left(\frac{k'}{\text{dn}^{2}\,4s\omega}\right)^{2}$$ and so $$l' = \dfrac{k'^{p}}{{\displaystyle\prod_{s = 1}^{(p - 1)/2}\text{dn}^{4}\,4s\omega}}$$ Putting $x = \text{sn}\,u$ in the expressions for $\sqrt{1 - y^{2}}, \sqrt{1 - l^{2}y^{2}}$ we get the following results: $$\begin{align}\text{cn}\left(\frac{u}{M}, l\right) &= \text{cn}\,u\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}(K - 4s\omega)}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,4s\omega}\notag\\ \text{dn}\left(\frac{u}{M}, l\right) &= \text{dn}\,u\prod_{s = 1}^{(p - 1)/2}\frac{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}(K - 4s\omega)}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,4s\omega}\notag\end{align}$$ The above formulas remain unchanged if $M$ is replaced by $-M$ and hence in the above formulas we can keep $$M = \prod_{s = 1}^{(p - 1)/2}\left(\frac{\text{sn}(K - 4s\omega)}{\text{sn}\,4s\omega}\right)^{2}$$ Now the values of $\text{sn}(u + 4\omega), \text{sn}(u + 8\omega), \ldots, \text{sn}(u + 2(p - 1)\omega)$ are the same as the values $\text{sn}(u + 2\omega), \text{sn}(u + 4\omega), \ldots, \text{sn}(u + (p - 1)\omega)$ except for the order of terms and sign of each term. It is best to check this by putting some actual odd value of $p$ (like 3, 5 etc). But upon squaring these values we see that the problem of sign no longer exists and therefore in each of the above relations we can actually replace $4s$ by $2s$ (by putting $u = 0$ and $u = K$ and noting that $\text{sn}(K + \alpha) = \text{sn}(K - \alpha)$. Thus we arrive at the following: $$\begin{align}\text{sn}\left(\frac{u}{M}, l\right) &= \frac{\text{sn}\,u}{M}\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}\,2s\omega}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,2s\omega}\notag\\ \text{cn}\left(\frac{u}{M}, l\right) &= \text{cn}\,u\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}(K - 2s\omega)}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,2s\omega}\notag\\ \text{dn}\left(\frac{u}{M}, l\right) &= \text{dn}\,u\prod_{s = 1}^{(p - 1)/2}\frac{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}(K - 2s\omega)}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,2s\omega}\notag\\ M &= \prod_{s = 1}^{(p - 1)/2}\left(\frac{\text{sn}(K - 2s\omega)}{\text{sn}\,2s\omega}\right)^{2}\notag\\ l &= k^{p}\prod_{s = 1}^{(p - 1)/2}\text{sn}^{4}(K - 2s\omega)\notag\\ l' &= \dfrac{k'^{p}}{{\displaystyle\prod_{s = 1}^{(p - 1)/2}\text{dn}^{4}\,2s\omega}}\notag\end{align}$$ In the development of these formulas so far we have assumed that $\omega = (mK + m'iK')/p$ where $m, m'$ are relatively prime to each other, but we have not given any specific values to $m, m'$. It turns that by giving different values to $m$ and $m'$ we arrive at different transformations. Out of all these transformations two are real i.e. the relation between $x$ and $y$ is expressed using real numbers as coefficients. We are going to study these two transformations next.

Jacobi's First Real Transformation

By choosing $m = 1, m' = 0$ we get $\omega = K/p$ which is a real number and from this we instantly see that all the numbers involved in the formulas mentioned above are real numbers. The corresponding formulas are as follows: $$\begin{align}\text{sn}\left(\frac{u}{M}, l\right) &= \frac{\text{sn}\,u}{M}\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}\,\dfrac{2sK}{p}}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,\dfrac{2sK}{p}}\tag{1}\\ \text{cn}\left(\frac{u}{M}, l\right) &= \text{cn}\,u\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}\left(K - \dfrac{2sK}{p}\right)}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,\dfrac{2sK}{p}}\tag{2}\\ \text{dn}\left(\frac{u}{M}, l\right) &= \text{dn}\,u\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\left(K - \dfrac{2sK}{p}\right)}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,\dfrac{2sK}{p}}\tag{3}\\ M &= \prod_{s = 1}^{(p - 1)/2}\left(\dfrac{\text{sn}\left(K - \dfrac{2sK}{p}\right)}{\text{sn}\,\dfrac{2sK}{p}}\right)^{2}\tag{4}\\ l &= k^{p}\prod_{s = 1}^{(p - 1)/2}\text{sn}^{4}\left(K - \frac{2sK}{p}\right)\tag{5}\\ l' &= \dfrac{k'^{p}}{{\displaystyle\prod_{s = 1}^{(p - 1)/2}\text{dn}^{4}\,\frac{2sK}{p}}}\tag{6}\end{align}$$ It is easy to see that the least positive value of $u$ for which the right side of $(1)$ vanishes is $u = 2K/p$ and the left side of this equation vanishes when $u/M = 2L$. Equating these two values of $u$ we get the relation $K/pM = L$.

Jacobi's First Complementary Transformation

Next we need to recast the above formulas in another form by using the identities: $$\begin{align}\dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}\,v}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,v} &= -\frac{\text{sn}(u + v)\,\text{sn}(u - v)}{\text{sn}^{2}\,v}\notag\\ \dfrac{1 - \dfrac{\text{sn}^{2}\,u}{\text{sn}^{2}(K - v)}}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,v} &= \frac{\text{cn}(u + v)\,\text{cn}(u - v)}{\text{cn}^{2}\,v}\notag\\ \frac{1 - k^{2}\text{sn}^{2}\,u\,\text{sn}^{2}(K - v)}{1 - k^{2}\,\text{sn}^{2}\,u\,\text{sn}^{2}\,v} &= \frac{\text{dn}(u + v)\,\text{dn}(u - v)}{\text{dn}^{2}\,v}\notag\end{align}$$ On applying these identities to the formulas $(1)$ and using $(4)$ to replace $M$ we get $$\begin{align}\text{sn}\left(\frac{u}{M}, l\right) &= (-1)^{(p - 1)/2}\,\text{sn}\,u\prod_{s = 1}^{(p - 1)/2}\left(\dfrac{\text{sn}\,\dfrac{2sK}{p}}{\text{sn}\left(K - \dfrac{2sK}{p}\right)}\right)^{2}\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\prod_{s = 1}^{(p - 1)/2}\dfrac{\text{sn}\left(u + \dfrac{2sK}{p}\right)\,\text{sn}\left(u - \dfrac{2sK}{p}\right)}{\text{sn}^{2}\,\dfrac{2sK}{p}}\notag\end{align}$$ and using equation $(5)$ we get $$\text{sn}\left(\frac{u}{M}, l\right) = (-1)^{(p - 1)/2}\sqrt{\frac{k^{p}}{l}}\prod_{s = -(p - 1)/2}^{(p - 1)/2}\text{sn}\left(u + \frac{2sK}{p}\right)\tag{7}$$ and similarly $$\begin{align}\text{cn}\left(\frac{u}{M}, l\right) &= \sqrt{\frac{l'k^{p}}{lk'^{p}}}\prod_{s = -(p - 1)/2}^{(p - 1)/2}\text{cn}\left(u + \frac{2sK}{p}\right)\tag{8}\\ \text{dn}\left(\frac{u}{M}, l\right) &= \sqrt{\frac{l'}{k'^{p}}}\prod_{s = -(p - 1)/2}^{(p - 1)/2}\text{dn}\left(u + \frac{2sK}{p}\right)\tag{9}\end{align}$$ Dividing the equation $(7)$ by equation $(8)$ we get $$\text{sc}\left(\frac{u}{M}, l\right) = (-1)^{(p - 1)/2}\sqrt{\frac{k'^{p}}{l'}}\prod_{s = -(p - 1)/2}^{(p - 1)/2}\text{sc}\left(u + \frac{2sK}{p}\right)$$ Replacing $u$ by $iu$ in the above equation and noting that $\text{sc}(iu, k) = i\,\text{sn}(u, k')$ we get $$\begin{align}\text{sn}\left(\frac{u}{M}, l'\right) &= \sqrt{\frac{k'^{p}}{l'}}\prod_{s = -(p - 1)/2}^{(p - 1)/2}\text{sn}\left(u - \frac{2siK}{p}, k'\right)\notag\\ &= \sqrt{\frac{k'^{p}}{l'}}\,\text{sn}(u, k')\prod_{s = 1}^{(p - 1)/2}\text{sn}\left(u + \frac{2siK}{p}, k'\right)\,\text{sn}\left(u - \frac{2siK}{p}, k'\right)\notag\\ &= (-1)^{(p - 1)/2}\prod_{s = 1}^{(p - 1)/2}\text{sn}^{2}\left(\frac{2siK}{p}, k'\right)\,\text{sn}(u, k')\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\prod_{s = 1}^{(p - 1)/2}\dfrac{1 - \dfrac{\text{sn}^{2}(u, k')}{\text{sn}^{2}\left(\dfrac{2siK}{p}, k'\right)}}{1 - k'^{2}\,\text{sn}^{2}(u, k')\,\text{sn}^{2}\left(\dfrac{2siK}{p}, k'\right)}\notag\end{align}$$ The factor on the right side which is independent of $u$ must be $1/M$ as can be easily seen when we divide both sides by $\text{sn}(u, k')$ and take limits as $u \to 0$.

Thus we finally obtain $$\text{sn}\left(\frac{u}{M}, l'\right) = \frac{\text{sn}(u, k')}{M}\prod_{s = 1}^{(p - 1)/2}\dfrac{1 + \dfrac{\text{sn}^{2}(u, k')}{\text{sc}^{2}\left(\dfrac{2sK}{p}, k\right)}}{1 + k'^{2}\,\text{sn}^{2}(u, k')\,\text{sc}^{2}\left(\dfrac{2sK}{p}, k\right)}$$ The only factor on the right side which can vanish is $\text{sn}(u, k')$ and the smallest value for which this vanishes is $u = 2K'$. The left side vanishes when $u/M = 2L'$ and therefore upon equating these two values of $u$ we get $K'/M = L'$.

We have thus derived the two relations $K/pM = L$ and $K'/M = L'$ from which we obtain the fundamental relation $L'/L = pK'/K$. From this it also follows that the value of $l$ occurring in these equations is actually less than $k$.

In the next post we will describe one more transformation which is also real and it leads to a value of $l$ which is greater than $k$ (which will be denoted by $l_{1}$) and in this case $K'/K = pL'_{1}/L_{1}$.

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