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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,

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

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

and,

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

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

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 $x^2+y^2=1$.)


Questions:

  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?
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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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,

$$\big(\alpha(\tau)\,\alpha(3\tau)\big)^2+\big(\beta(\tau)\,\beta(3\tau)\big)^2=1\tag1\quad$$

as well as,

$$\left(\frac{a^2(2\tau)\,b(\tau)}{a(\tau)\,b^2(2\tau)}\right)^2+\big(2a(\tau)\,a(2\tau)\big)^2=1\quad\tag2$$

$$\left(\frac{c^2(\tau)\,d(2\tau)}{c(2\tau)\,d^2(\tau)}\right)^2+\big(2c(\tau)\,c(2\tau)\big)^2=1\quad\tag3$$

$$\quad\left(\frac{a^2(2\tau)}{a(\tau)}\right)^2-\big(2a(\tau)\,a(2\tau)\big)^2=\big(\sqrt3\,d(2\tau)\big)^2\tag4$$

$$\quad\left(\frac{c^2(\tau)}{c(2\tau)}\right)^2-\big(2c(\tau)\,c(2\tau)\big)^2=\big(\sqrt3\,b(\tau)\big)^2\tag5$$

$$\left(\frac{a^2(2\tau)}{a(\tau)}\right)^2+\left(\frac{2a(\tau)\,a(2\tau)}{b(\tau)}\right)^2=1\quad\tag6$$

$$\left(\frac{c^2(\tau)}{c(2\tau)}\right)^2+\left(\frac{2c(\tau)\,c(2\tau)}{d(2\tau)}\right)^2=1\quad\tag7$$

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.

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