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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