Modular equation for a quotient of eta functions

Question

In this article at page $4$ F. Beukers introduces a function $$y(\tau) = \frac{\eta^{8}(6\tau)\eta^{4}(\tau)}{\eta^{8}(2\tau)\eta^{4}(3\tau)}\tag{1}$$ where $\tau$ is a complex number with positive imaginary part and Dedekind's eta function $\eta(\tau)$ is defined by $$\eta(\tau) = e^{\pi i\tau/12}\prod_{n = 1}^{\infty}(1 - e^{2\pi in\tau})\tag{2}$$ Beukers then mentions that the function $y(\tau)$ satisfies the following equation $$y\Big(\frac{-1}{6\tau}\Big) = \frac{y(\tau) - 1/9}{y(\tau) - 1}\tag{3}$$

But I can't check that. I have already tried to use the functional equation for eta function $$\eta(-1/\tau)=\sqrt{-i\tau}\;\eta(\tau)\tag{4}$$ but this doesn't work. Has this something to do with Beukers remark "$y(\tau)$ generates the field of modular functions"? But why does $y(\tau)$ generate this field?

Thanks a lot in advance.

Answer 1

Since I am not so much familiar with the theory of modular forms, I am offering a solution which uses theory of theta functions and their link with elliptic integrals. With this approach the functional relation of $y(\tau)$ is transformed into a modular equation of degree $3$ which can be independently verified using the theory of modular equations.

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Let $\tau$ be a complex number with positive imaginary part so that $q = e^{2\pi i\tau}$ satisfies $|q| < 1$. The Dedekind $\eta$ function is defined as $$\eta(\tau) = e^{\pi i\tau/12}\prod_{n = 1}^{\infty}(1 - e^{2\pi in\tau}) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n}) = 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{1}$$ where $k, k', K$ associated with nome $q$. Further we have $$\eta(2\tau) = q^{1/12}\prod_{n = 1}^{\infty}(1 - q^{2n}) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{2}$$ Now let $l,l', L$ be associated with $q' = e^{6\pi i\tau} = q^{3}$. Then we have $$\eta(3\tau) = 2^{-1/6}\sqrt{\frac{2L}{\pi}}l^{1/12}l'^{1/3}\tag{3}$$ and $$\eta(6\tau) = 2^{-1/3}\sqrt{\frac{2L}{\pi}}(ll')^{1/6}\tag{4}$$ Beukers' paper talks about the function $$y(\tau) = \frac{\eta^{8}(6\tau)\eta^{4}(\tau)}{\eta^{8}(2\tau)\eta^{4}(3\tau)}\tag{5}$$ and mentions that it satisfies the following identity $$y(-1/6\tau) = \frac{y(\tau) - 1/9}{y(\tau) - 1}\tag{6}$$ Using the expression of $y(\tau)$ in terms of $\eta$ function and the functional equation $$\eta(-1/\tau) = \sqrt{-i\tau}\eta(\tau)\tag{7}$$ satisfied by the $\eta$ function we see that equation $(6)$ is equivalent to the following identity $$\frac{\eta^{8}(\tau)\eta^{4}(6\tau)}{\eta^{8}(3\tau)\eta^{4}(2\tau)} = \frac{9\eta^{8}(6\tau)\eta^{4}(\tau) - \eta^{8}(2\tau)\eta^{4}(3\tau)}{\eta^{8}(6\tau)\eta^{4}(\tau) - \eta^{8}(2\tau)\eta^{4}(3\tau)}\tag{8}$$ or $$\eta^{12}(\tau)\eta^{12}(6\tau) - \eta^{8}(\tau)\eta^{8}(2\tau)\eta^{4}(3\tau)\eta^{4}(6\tau) = 9\eta^{4}(\tau)\eta^{4}(2\tau)\eta^{8}(3\tau)\eta^{8}(6\tau) - \eta^{12}(2\tau)\eta^{12}(3\tau)$$ and using equations $(1)-(4)$ we can see that this is equivalent to $$\left(\frac{4KL}{\pi^{2}}\right)^{2}k'^{2}l - \left(\frac{2K}{\pi}\right)^{4}kk'^{2} = 9\left(\frac{2L}{\pi}\right)^{4}ll'^{2} - \left(\frac{4KL}{\pi^{2}}\right)^{2}kl'^{2}$$ Using $m = K/L$ for multiplier we see that the above is equivalent to $$\frac{k'^{2}kl}{m^{2}} - k^{2}k'^{2} = \frac{9kll'^{2}}{m^{4}} - \frac{k^{2}l'^{2}}{m^{2}}\tag{9}$$ Now we know from the theory of modular equations of degree $3$ that $$k^{2} = \frac{(m - 1)(3 + m)^{3}}{16m^{3}},\,l^{2} = \frac{(m - 1)^{3}(3 + m)}{16m}$$ and $$k'^{2} = \frac{(m + 1)(3 - m)^{3}}{16m^{3}},\,l'^{2} = \frac{(m + 1)^{3}(3 - m)}{16m}$$ so that $$kl = \frac{(m - 1)^{2}(3 + m)^{2}}{16m^{2}}$$ Putting the values of $k, l, k', l'$ in terms of $m$ we see (after some simple cancellations of factors on both sides of the equation $(9)$) that equation $(9)$ holds if we can establish $$(3 - m)^{2}(m - 1) - m(3 - m)^{2}(3 + m) = 9(m - 1)(m + 1)^{2} - m(3 + m)(m + 1)^{2}\tag{10}$$ Since each side is a polynomial of degree $4$ in $m$ it follows that the relation $(10)$ holds identically for all values of $m$ if it holds for at least $5$ distinct values of $m$. We can verify very easily that it holds for $m = 0, 1, -1, 3, -3$ and hence equation $(10)$ holds. Thus we have proved that $(9)$ also holds.

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The approach given above is more of a verification of the identity $(6)$ which Beukers mentions in his paper. It is desirable to seek a proof based on some identity relating eta functions of argument $\tau, 2\tau, 3\tau$ or perhaps a proof based on theory of modular forms which shows directly that $y(-1/6\tau)$ is a rational function of $y(\tau)$.

Answer 2

There is a simple proof using two eta product identities. If $$ y_1(\tau) := \eta(2\tau)\eta^{9}(6\tau)\eta^{4}(\tau), \quad y_2(\tau) := \eta^{9}(2\tau)\eta^{4}(3\tau)\eta(6\tau), $$ then $\, y(\tau) = y_1(\tau)/y_2(\tau), \,$ and

$$ y_2(\tau) - 9y_1(\tau) = \eta(\tau)^9\eta(3\tau)\eta(6\tau)^4, \tag{1} $$

$$ y_2(\tau) - y_1(\tau) = \eta(\tau)\eta(2\tau)^4\eta(3\tau)^9, \tag{2} $$

$$ y_3(\tau) := \frac{y_2(\tau) - 9y_1(\tau)}{y_2(\tau) - y_1(\tau)} = 9\frac{y(\tau) - 1/9}{y(\tau) - 1}, \tag{3} $$

$$ y_3(\tau) = 9 \frac{\eta(\tau)^8 \eta(6\tau)^4} {\eta(3\tau)^8 \eta(2\tau)^4} = y\Big(\frac{-1}{6\tau}\Big). \tag{4}$$

Equation $(1)$ is eta product identity $t_{6,13,30}$ multiplied by $\,\eta(6\tau)\,$ and equation $(2)$ is eta product identity $t_{6,13,34}$ multiplied by $\,\eta(2\tau).\,$ Both identities have many different forms and references to proofs were listed in their entry in my Dedekind eta product identities website. The Wikipedia article Dedekind eta function links to the Wayback machine where my website is archived.

Answer 3

Identity (8) corresponds to an identity of Ramanujan proved by Berndt. "Ramanujan Notebooks", Part IV, p. 204, Entry 51.