TITLE: Modular equation for a quotient of eta functions
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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
$$ y_2(\tau) - 9y_1(\tau) = \eta(\tau)^9\eta(3\tau)\eta(6\tau)^4, \tag{1} $$
$$ y_2(\tau) -  y_1(\tau) = \eta(\tau)\eta(2\tau)^4\eta(3\tau)^9, \tag{2} $$
$$ y_3(\tau) := \frac{y_2(\tau) - 9y_1(\tau)}{y_2(\tau) -  y_1(\tau)}
     = 9\frac{y(\tau) - 1/9}{y(\tau) - 1}, \tag{3} $$
$$ y_3(\tau) =  9 \frac{\eta(\tau)^8 \eta(6\tau)^4}{\eta(2\tau)^4 \eta(3\tau)^8}
     = y\Big(\frac{-1}{6\tau}\Big). \tag{4}$$
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 are listed in
their entry in my Dedekind eta product identities website.