# Elementary Approach to Modular Equations: Jacobi's Transformation Theory 3

### Analytic Approach to Transformation Theory

Jacobi understood that the algebraic approach for obtaining modular equations could not be applied easily in case of higher degrees. Hence he followed an analytic approach. The idea he used was to express the relation $y = xN(1, x^{2})/D(1, x^{2})$ in a form where each of $N$ and $D$ appears as a product of various factors. Essentially he examined the roots of $N, D$ and expressed them in form of a product where each factor corresponds to a given root. This approach was very useful for Jacobi as he used this relation to finally develop the theory of Theta Functions and their relation to elliptic functions. In fact the entire Fundamenta Nova is split into two sections, the first section dealing with transformation theory and the second section dealing with the expansion of elliptic functions into infinite series and products (which is basically the theory of theta functions).

The presentation here is based on the Calyley's An Elementary Treatise on Elliptic Functions and our goal here will be to establish the existence of transformation of odd prime degree and to establish the fundamental relation $L'/L = pK'/K$ which is induced by these transformations. We will not explore the development of theta functions based on this transformation theory as done in Jacobi's or Cayley's books.

To begin with lets fix the notation which is almost similar to that used by Jacobi/Cayley except that we use $p$ in place of $n$ and $l$ in place of $\lambda$. Let's assume that $m$ and $m'$ are some fixed integers relatively prime to each other not both zero and let $$\omega = \frac{mK + m'iK'}{p}$$ The transformation is given by $y = U/V$ and the resulting equations \begin{align} V + U &= (1 + x)A^{2},\,\,\,\, V - U = (1 - x)B^{2}\notag\\ 1 + y &= (1 + x)A^{2}/V,\,\,\,\, 1 - y = (1 - x)B^{2}/V\notag\\ V + lU &= (1 + kx)C^{2},\,\,\,\, V - lU = (1 - kx)D^{2}\notag\\ 1 + ly &= (1 + kx)C^{2}/V,\,\,\,\, 1 - ly = (1 - kx)D^{2}/V\notag\end{align} The polynomials $U, V$ are given by \begin{align} U &= \frac{x}{M}\left(1 - \frac{x^{2}}{\text{sn}^{2}(4\omega, k)}\right)\left(1 - \frac{x^{2}}{\text{sn}^{2}(8\omega, k)}\right)\cdots\left(1 - \frac{x^{2}}{\text{sn}^{2}(2(p - 1)\omega, k)}\right)\notag\\ &= \frac{x}{M}\prod_{s = 1}^{(p - 1)/2}\left(1 - \frac{x^{2}}{\text{sn}^{2}(4s\omega, k)}\right)\notag\\ V &= (1 - k^{2}x^{2}\text{sn}^{2}(4\omega, k))(1 - k^{2}x^{2}\text{sn}^{2}(8\omega, k))\cdots(1 - k^{2}x^{2}\text{sn}^{2}(2(p - 1)\omega, k))\notag\\ &= \prod_{s = 1}^{(p - 1)/2}(1 - k^{2}x^{2}\text{sn}^{2}(4s\omega, k))\notag\end{align} and the polynomials $A, B, C, D$ are given by \begin{align} A &= \left(1 + \frac{x}{\text{sn}(K - 4\omega, k)}\right)\left(1 + \frac{x}{\text{sn}(K - 8\omega, k)}\right)\cdots\left(1 + \frac{x}{\text{sn}(K - 2(p - 1)\omega, k)}\right)\notag\\ &= \prod_{s = 1}^{(p - 1)/2}\left(1 + \frac{x}{\text{sn}(K - 4s\omega, k)}\right)\notag\\ B &= \left(1 - \frac{x}{\text{sn}(K - 4\omega, k)}\right)\left(1 - \frac{x}{\text{sn}(K - 8\omega, k)}\right)\cdots\left(1 - \frac{x}{\text{sn}(K - 2(p - 1)\omega, k)}\right)\notag\\ &= \prod_{s = 1}^{(p - 1)/2}\left(1 - \frac{x}{\text{sn}(K - 4s\omega, k)}\right)\notag\\ C &= (1 + kx\,\text{sn}(K - 4\omega, k))(1 + kx\,\text{sn}(K - 8\omega, k))\cdots(1 + kx\,\text{sn}(K - 2(p - 1)\omega, k))\notag\\ &= \prod_{s = 1}^{(p - 1)/2}(1 + kx\,\text{sn}(K - 4s\omega, k))\notag\\ D &= (1 - kx\,\text{sn}(K - 4\omega, k))(1 - kx\,\text{sn}(K - 8\omega, k))\cdots(1 - kx\,\text{sn}(K - 2(p - 1)\omega, k))\notag\\ &= \prod_{s = 1}^{(p - 1)/2}(1 - kx\,\text{sn}(K - 4s\omega, k))\notag\end{align} The invariance of relation between $x$ and $y$ under the transformation $(x, y) \to (1/kx, 1/ly)$ is satisfied if we have \begin{align}l &= k^{p}\{\text{sn}(K - 4\omega, k)\,\text{sn}(K - 8\omega, k)\cdots\,\text{sn}(K - 2(p -1)\omega, k)\}^{4}\notag\\ &= k^{p}\prod_{s = 1}^{(p - 1)/2}(\text{sn}(K - 4s\omega, k))^{4}\notag\\ M &= (-1)^{(p -1)/2}\left\{\frac{\text{sn}(K - 4\omega, k)}{\text{sn}(4\omega, k)}\,\frac{\text{sn}(K - 8\omega, k)}{\text{sn}(8\omega, k)}\,\cdots\,\frac{\text{sn}(K - 2(p - 1)\omega, k)}{\text{sn}(2(p - 1)\omega, k)}\right\}^{2}\notag\\ &= (-1)^{(p - 1)/2}\prod_{s = 1}^{(p - 1)/2}\left(\frac{\text{sn}(K - 4s\omega, k)}{\text{sn}(4s\omega, k)}\right)^{2}\notag\end{align} In order to prove that the above mentioned transformation indeed has all the properties desired takes some reasonable amount of algebraic manipulations and some analysis of the various equations involved. In what follows elliptic functions will be used heavily and therefore we will drop the modulus $k$. If any other modulus is used for these elliptic elliptic functions it will written explicitly.

We start with the equation $$\frac{1 - y}{1 + y} = \frac{1 - x}{1 + x}\prod_{s = 1}^{(p - 1)/2}\left(\dfrac{1 - \dfrac{x}{\text{sn}(K - 4s\omega)}}{1 + \dfrac{x}{\text{sn}(K - 4s\omega)}}\right)^{2}$$ and will show that this leads to all the other equations. Moreover from the form of the equations given earlier it follows that $$\frac{Mdy}{\sqrt{(1 - y^{2})(1 - l^{2}y^{2})}} = \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}$$ Upon solving for $y$ we will obtain an expression $y = xN(1, x^{2})/D(1, x^{2})$, where $N, D$ are homogeneous polynomials of degree $(p - 1)/2$ (this is because $x, y$ change their sign together and thus $y/x$ is a rational and even function).

Thus if we are able to show that $y$ vanishes for $x = \pm\,\text{sn}\,4t\omega, t = 0, 1, 2,\ldots, (p - 1)/2$and that $y \to \pm\infty$ as $x \to \pm 1/(k\,\text{sn}\,4t\omega), t =1, 2, \ldots, (p - 1)/2$ then it will be obvious that $y$ has to be of the form: $$y = \frac{x}{C}\prod_{s = 1}^{(p - 1)/2}\left(\dfrac{1 - \dfrac{x^{2}}{\text{sn}^{2}4s\omega}}{1 - k^{2}x^{2}\text{sn}^{2}4s\omega}\right)$$ Also from the relation between $x, y$ it is clear that $y = 1$ when $x = 1$ and therefore we must have \begin{align}C &= \prod_{s = 1}^{(p - 1)/2}\left(\dfrac{1 - \dfrac{1}{\text{sn}^{2}4s\omega}}{1 - k^{2}\text{sn}^{2}4s\omega}\right) = (-1)^{(p - 1)/2}\prod_{s = 1}^{(p - 1)/2}\left(\frac{\text{cn}\,4s\omega}{\text{sn}\,4s\omega\,\text{dn}\,4s\omega}\right)^{2}\notag\\ &= (-1)^{(p - 1)/2}\prod_{s = 1}^{(p - 1)/2}\left(\frac{\text{sn}(K - 4s\omega)}{\text{sn}\,4s\omega}\right)^{2} = M\notag\end{align} With this expression of $y$ we at once see that $y = U/V$ where $U, V$ are as mentioned above. Also if assume the invariance of this relation under the transformation $(x, y) \to (1/kx, 1/ly)$ we at once arrive at the value of $l$ described above. Again from equation $(1 - y)/(1 + y) = ((1 - x)B^{2})/((1 + x)A^{2})$ we obtain $$y = \frac{U}{V} = \frac{(1 + x)A^{2} - (1 - x)B^{2}}{(1 + x)A^{2} + (1 - x)B^{2}}$$ Looking at the degrees of $V$ and $(1 + x)A^{2} + (1 - x)B^{2}$ we can see that $V$ must be a constant multiple of $(1 + x)A^{2} + (1 - x)B^{2}$ and on putting $x = 0$ we see that $V = (1/2)((1 + x)A^{2} + (1 - x)B^{2})$ and so $U = (1/2)((1 + x)A^{2} - (1 - x)B^{2})$ and hence it follows that $1 + y = (V + U)/V = ((1 + x)A^{2})/V$ and $1 - y = (V - U)/V = ((1 - x)B^{2})/V$. Thus the expressions of $1 - y, 1 + y$ are also verified. Using the invariance under $(x, y) \to (1/kx, 1/ly)$ we can then verify the expressions for $1 - ly, 1 + ly$.

Hence all the above expressions relating $x$ and $y$ are verified provided we establish that zeros and poles (points where a function tends to $\pm\infty$) of $y$ are as described above. To that end we put $x = \text{sn}\,u$ in the expession for $(1 - y)/(1 + y)$ and obtain $$\frac{1 - y}{1 + y} = \frac{1 - \text{sn}\,u}{1 + \text{sn}\,u}\prod_{s = 1}^{(p - 1)/2}\left(\dfrac{1 - \dfrac{\text{sn}\,u}{\text{sn}(K - 4s\omega)}}{1 + \dfrac{\text{sn}\,u}{\text{sn}(K - 4s\omega)}}\right)^{2}$$ Now we require some identities related to the elliptic functions which can be verified at once using addition formulas or otherwise: $$\{(1 + \text{sn}(u + v)\}\{1 + \text{sn}(u - v)\} = \dfrac{\text{cn}^{2}v\left(1 + \dfrac{\text{sn}\,u}{\text{sn}(K - v)}\right)^{2}}{1 - k^{2}\text{sn}^{2}u\,\text{sn}^{2}v}$$ $$\{(1 - \text{sn}(u + v)\}\{1 - \text{sn}(u - v)\} = \dfrac{\text{cn}^{2}v\left(1 - \dfrac{\text{sn}\,u}{\text{sn}(K - v)}\right)^{2}}{1 - k^{2}\text{sn}^{2}u\,\text{sn}^{2}v}$$ and therefore $$\frac{\{(1 - \text{sn}(u + v)\}\{1 - \text{sn}(u - v)\}}{\{(1 + \text{sn}(u + v)\}\{1 + \text{sn}(u - v)\}} = \left(\dfrac{1 - \dfrac{\text{sn}\,u}{\text{sn}(K - v)}}{1 + \dfrac{\text{sn}\,u}{\text{sn}(K - v)}}\right)^{2}$$ In the above relation we put $v = 4\omega, 8\omega, \ldots, 2(p - 1)\omega$ successively and multiply the resulting equations and further multiply both sides by $(1 - \text{sn}\,u)/(1 + \text{sn}\,u)$. Also while doing so we need to observe that $\text{sn}(u - 4s\omega) = \text{sn}(u + 4(p - s)\omega)$. Thus we obtain the following: $$\frac{1 - y}{1 + y} = \prod_{s = 0}^{p - 1}\frac{1 - \text{sn}(u + 4s\omega)}{1 + \text{sn}(u + 4s\omega)}$$ From the above relation it is easy to see that the $(1 - y)/(1 + y)$ and hence $y$ is invariant when $u$ is replaced $u + 4\omega$ (in doing so each factor is changed into next factor in the product and the last factor is changed into the first so that the product itself remains invariant). If we put $u = 0$ we have $x = 0$ and hence $y = 0$. Therefore $y = 0$ whenenver $u = 0, 4\omega, 8\omega, \ldots, 4(p - 1)\omega$ i.e. whenever $x = \text{sn}\,4t\omega, t = 0, 1, 2, \ldots, (p - 1)$. Since $\text{sn}\,4(p - t)\omega = -\text{sn}\,4t\omega$ it follows that $y = 0$ whenever $x = 0$ or $x = \pm\,\text{sn}\,4t\omega, t = 0, 1, 2, \ldots, (p - 1)/2$. Thus the zeroes of $y$ are verified.

It is easily seen that $u = iK'$ is a pole for $x = \text{sn}\,u$ and hence a pole for $y$ (clearly $y \to \infty$ as $x \to \infty$). Thus $y \to \infty$ whenever $x \to \text{sn}(iK' + 4t\omega), t = 1, 2, \ldots, (p - 1)$ i.e whenever $x \to \text{sn}(iK' \pm 4t\omega) = \pm 1/(k\,\text{sn}\,4t\omega), t = 1, 2, \ldots, (p - 1)/2$ and so the position of poles of $y$ are also verified.

Thus the tranformations of order $p$ described in the beginning of the post using complicated product expansions containing elliptic functions are verified. We will discuss the repercussions of these transformations in the next post.

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1. please the pdf file is broken thank you

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2. PDF link fixed.