Given the Dedekind eta function $\eta(\tau)$, define,

$$\alpha(\tau) =\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}$$

$$\beta(\tau) =\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\quad\;$$

$$\quad\gamma(\tau) =\frac{\alpha(\tau)}{\beta(\tau)} =\frac{\sqrt2\,\eta(4\tau)}{\eta(\tau)}$$

Note that $\alpha^8(\tau)=\lambda(2\tau)$ with modular lambda function $\lambda(\tau)$. We have the well-known,

$$\alpha^8(\tau)+\beta^8(\tau) = 1\tag1$$

as well as the nice,

$$\frac{1}{\beta^2(\tau)}-\beta^2(\tau) =\left(2^{1/4}\,\gamma(2\tau)\right)^4\tag2$$

$$\frac{1}{\alpha^2(2\tau)}-\alpha^2(2\tau) =\left(\frac{2^{1/4}}{\gamma(\tau)}\right)^4\tag3$$


$$\frac{1}{\alpha^2(2\tau)}+\alpha^2(2\tau) =\left(\frac{2^{1/4}}{\alpha(\tau)}\right)^4\tag4$$

$$\frac{1}{\beta^2(\tau)}+\beta^2(\tau) =\left(\frac{2^{1/4}}{\beta(2\tau)}\right)^4\tag5$$

As eta functions (not as quotients $\alpha$ and $\beta$), these 5 are in Somos' database (as t4_24_48, t8_12_24, t8_12_48, t8_18_60a, t8_18_60b). After some algebraic manipulation, I realized these can be expressed in more aesthetic forms. (Of course, it is easy to transform these to the form $x^2+y^2=1$.)


  1. For the next step, can we express $$\frac{1}{\beta^4(\tau)}+\beta^4(\tau) =t_1^2$$ $$\frac{1}{\alpha^4(2\tau)}+\alpha^4(2\tau) =t_2^2$$ where the $t_i$ are eta quotients?
  2. Are there are other eta quotient parameterizations strictly of form, $$\left(m_1\prod\eta(a_1\tau)^{c_1}\,\eta(a_2\tau)^{c_2}\dots\right)^2 + \left(m_2\prod\eta(b_1\tau)^{d_1}\,\eta(b_2\tau)^{d_2}\dots\right)^2 = 1$$ where $a_i, b_i, c_i, d_i$ are integers (the exponents $c_i,d_i$ may be negative) and $m_i$ are algebraic similar to the above 5?
  • $\begingroup$ After some tinkering, the best I could do was, $$\frac{1}{\alpha^4(2\tau)}+\alpha^4(2\tau) =\frac{4}{\alpha^8(\tau)}-2$$ $$\frac{1}{\beta^4(\tau)}+\beta^4(\tau) =\frac{4}{\beta^8(2\tau)}-2$$ which just factors into the known identities $(4),(5)$. Sigh. $\endgroup$ – Tito Piezas III Mar 7 at 14:35
  • $\begingroup$ If $f$ is weight $0$ modular for $\Gamma$ and $f,1-f$ have no odd order zeros or poles on the upper half-plane then won't $f^{1/2},(1-f)^{1/2}$ be modular for some index $|4$ subgroup ? $\endgroup$ – reuns Mar 7 at 20:34

For your question $2$ all that is needed is a quick search of 'eta07.gp' for three terms identities with even rank and with all even exponents. I turned up $t_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t_{12,12,40},\; t_{12,12,56},\; t_{12,24,72},\; t_{12,24,90},\; t_{12,24,120},\; t_{12,24,126}.$$ For the first, the identity is equivalent to $$ \left(\sqrt{\frac14}\,\frac{u_1^3\,u_{12}}{u_3\,u_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u_2^4\,u_6^2}{u_1\,u_3\,u_4^4}\right)^2 = 1 $$ where $u_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.

Note the two transforms. First, given an eta identity, replacing $q$ with $-q$ gives an eta identity. Second, the same with replacing $q^d$ with $q^{N/d}$ where $N$ is the level. By iterating these two transforms we get an eta identity cluster. There can be up to $12$ identities in a cluster.

  • $\begingroup$ Beautiful! So we have for level $4,8,12$. Is there for level $16,20$? $\endgroup$ – Tito Piezas III Mar 8 at 3:43
  • $\begingroup$ The identity t12_12_48b = $1*u1^4*u3^4*u4^2*u12^2 + 4*q*u1^2*u3^2*u4^4*u12^4 - 1*u2^6*u6^6$ works as well since it has only even powers. $\endgroup$ – Tito Piezas III Mar 8 at 4:56
  • $\begingroup$ @TitoPiezasIII Great! I somehow missed that one. I should have had an automated search function written by now. Give me a few hours. $\endgroup$ – Somos Mar 8 at 5:16

Courtesy of Somos' answer, we can find more eta quotient parameterizations to $x^2+y^2 = 1$. Define,

$$a(\tau) = \frac{\eta^3(\tau)\,\eta(6\tau)}{2\,\eta^3(2\tau)\,\eta(3\tau)},\quad\quad b(\tau) = \frac{\eta^2(\tau)\,\eta(6\tau)}{\sqrt3\,\eta^2(3\tau)\,\eta(2\tau)}$$

$$c(\tau) = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)},\quad\quad d(\tau) = \frac{\eta(\tau)\,\eta^2(6\tau)}{\eta(3\tau)\,\eta^2(2\tau)}\quad$$

Note that $c(\tau)$ is the cubic continued fraction and the four obey,

$$\frac{-1}{a^3(\tau)} +\frac1{b^4(\tau)}= 1,\quad\quad \frac{-1}{c^3(\tau)} +\frac1{d^4(\tau)}= 1$$

For the first identity, recall $\alpha(\tau)$ and $\beta(\tau)$ from the original post. Then we have,


as well as,







The first one being $t_{12,12,48b}$ while the rest are the 6 mentioned by Somos.

P.S. This is a revised version of the original answer since, as I suspected, they could be expressed in a more consistent form.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.