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.

  • $\begingroup$ It is better to write at least some details about equation $(1)$. Also mention the definition of $\eta$ function and $y(\tau)$. BTW I did some calculation and it appears that this is related to modular equation between $\eta(\tau), \eta(2t), \eta(3t), \eta(6t)$. A proof would not be easy. I need to check some references to see if it can be done in simple manner. $\endgroup$
    – Paramanand Singh
    Jun 5, 2016 at 17:10
  • $\begingroup$ @ParamanandSingh: It seems this functional equation belongs to a family. Kindly see this MO post $\endgroup$ Feb 4, 2017 at 11:27

3 Answers 3


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 $y(\tau)$.

  • $\begingroup$ @R.A.: It appears that you have been so motivated by "easy to show". I think Beukers' means that it is easy if one has good knowledge of theory of modular forms. I actually don't know much about modular forms, rather I am able to link various modular functions like $\eta$ with elliptic integrals $K$ and modulus $k, k'$ and transform identities between eta functions into identities between $k, k', K$. All this is part of a difficult theory of "elliptic integrals/functions and theta functions" and some part of it was later developed into theory of modular forms. $\endgroup$
    – Paramanand Singh
    Jun 6, 2016 at 4:43
  • $\begingroup$ @R.A.: The meaning of "$y(\tau)$ generates a field" is something very different. Beukers wants to say that the field of rational expression in $y(\tau)$ also contains $y(-1/6\tau)$ and also gives an expression of $y(-1/6\tau)$ in terms of $y(\tau)$. But why this is the case needs a proof and the proof is non-trivial either by theory of modular forms (which I don't know) or by theory of theta functions (which I have given in my answer). Please don't get carried away by "easy to show" phrase written by Beukers. $\endgroup$
    – Paramanand Singh
    Jun 6, 2016 at 7:16
  • $\begingroup$ @R.A.: Moreover if $x = y(\tau)$ then a rational expression in $x$ means a function of type $f(x)/g(x)$ where $f, g$ are polynomials. Thus he says that $y(-1/6\tau) = f(y(\tau))/g(y(\tau))$ for some some polynomials $f, g$. And coefficients of these polynomials are algebraic numbers. $\endgroup$
    – Paramanand Singh
    Jun 6, 2016 at 7:18
  • $\begingroup$ @R.A.: It is not easy to find exact polynomials $f, g$ (and not possible via Fourier analysis). Beukers says that in this particular case $f(x) = x - (1/9)$ and $g(x) = x - 1$ but to get these expressions for $f, g$ is not as easy as Beukers' writes (definitely not easy without theory of modular forms). I suggest you study some theory of modular forms. And don't worry if you don't understand my solution. It also belongs to a difficult theory of elliptic integrals/functions and theta functions and not everyone is familiar with it. $\endgroup$
    – Paramanand Singh
    Jun 6, 2016 at 7:20
  • $\begingroup$ If $\tau$ is interpreted as the elliptic period ratio, the rightmost sides of $(1)$ and $(2)$ give $\eta\left(\frac{\tau}{2}\right)$ and $\eta(\tau)$ respectively. (The discrepancy may be due to the conflicting definitions for $q$ in use.) This does not invalidate the following reasoning because $\tau$ could be replaced with $2\tau$ everywhere. Nice proof. (I upvoted long ago, even before checking the details.) $\endgroup$
    – ccorn
    Apr 26, 2018 at 13:27

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.


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

  • $\begingroup$ The eta-product identity is $q_{6,24,60a}$ in my collection. The Ramanujan identity it is equivalent to is $(PQ)+9q/(PQ) = (Q/P)^3+q/(Q/P)^3$ where $P=(f(-q)/f(-q^3))^2, Q=(f(-q^2)/f(-q^6))^2$ and $f(-q)$ is a Ramanujan theta function, essentially $\eta(q)$ without the $q^{1/24}$ factor. $\endgroup$
    – Somos
    Oct 20, 2017 at 0:47

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