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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.

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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.

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.

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 $y(\tau)$.

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

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