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Define $\color{blue}{q=e^{2\pi i \tau}}$ and Dedekind eta function $\eta(\tau)$. Note: I found these relations empirically, but their consistent forms suggest they can be rigorously proven.

I. $p=2$

We have,

$$\left(\frac{8}{\alpha^8+8}\right)^2+\left(\frac{\beta^8}{\beta^8+32}\right)^2=1\tag1$$ $$\alpha=\frac{\eta(\tau/2)}{\eta(2\tau)},\quad\beta=\frac{\eta(\tau)}{\eta(4\tau)}$$ If $\tau=\sqrt{-n}$, then, $$\frac{_2F_1\Big(\tfrac14,\tfrac34;1;\,1-\big(\tfrac{8}{\alpha^2+8}\big)^2\Big)}{_2F_1\Big(\tfrac14,\tfrac34;1;\,\big(\tfrac{8}{\alpha^8+8}\big)^2\Big)}=\color{red}{\sqrt{2n}}$$

II. $p=3$

We have,

$$\left(\frac{3}{\gamma^3+3}\right)^3+\left(\frac{\delta^3}{\delta^3+9}\right)^3=1\tag2$$ $$\gamma=\frac{\eta(\tau/3)}{\eta(3\tau)},\quad\delta=\frac{\eta(\tau)}{\eta(9\tau)}$$ If $\tau=\sqrt{-n}$, then, $$\frac{_2F_1\Big(\tfrac13,\tfrac23;1;\,1-\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}{_2F_1\Big(\tfrac13,\tfrac23;1;\,\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}=\color{red}{\sqrt{3n}}$$

Note: The identity, $$\left(\frac{x^3y + y}{x y^3 + x}\right)^3+\left(\frac{x^3 - y^3}{x y^3 + x}\right)^3=1,\quad \text{if}\; x^3+y^3=1$$ should guarantee infinitely many eta parametrizations to $(2)$. Note also that, $$\gamma^3+3=4C^2(q)+\frac1{C(q)}\\ \delta^3+3=4C^2(q^3)+\frac1{C(q^3)}$$ with cubic continued fraction, $$C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}=\cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}$$

$\color{blue}{Update:}$ I just realized that the cubic eta identity $(2)$ is equivalent to the Borweins' cubic theta identity, $$\frac{c^3(q)}{a^3(q)}+\frac{b^3(q)}{a^3(q)} = 1$$ where, $$a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$ $$b(q) = \tfrac{1}{2}\big(3a(q^3)-a(q)\big)$$ $$c(q) = \tfrac{1}{2}\big(a(q^{1/3})-a(q)\big)$$ from Ramanujan's Notebooks Vol. V, p.93.

III. $p=4,8$

$$\left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4+\left(\frac{\vartheta_4(0,q)}{\vartheta_3(0,q)}\right)^4 = 1\tag{3a}$$ with Jacobi theta function $\vartheta_n(0,q)$. If $q=e^{2\pi i \tau}$, then equivalently,

$$\left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8+\left(\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\right)^8 = 1\tag{3b}$$

hence the addends of $(3a)$ and $(3b)$ are equal. Expressed as, $$U^8(\tau)+V^8(\tau) =1$$

If $\tau=\sqrt{-n}$, then,

$$\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-U^8(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,U^8(\tau)\big)}=\color{red}{\sqrt{4n}}$$

This has a beautiful octic continued fraction studied by Ramanujan, $$U(\tau)=\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}= \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$ and can solve the general quintic.


IV. Question

Q: Are there analogous eta quotients to parameterize the Fermat quintic $x^5+y^5=1$? And can we also use the Rogers-Ramanujan cfrac to do so?

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    $\begingroup$ There's even a parametrization for $x^4+y^4=1$ and $x^8+y^8=1$ (or at least $ax^8 + by^8 = 1$ for some rational $a,b$). Also sixth powers, if memory serves. But fifth powers, no. Not with eta products/quotients, anyhow. $\endgroup$ – Noam D. Elkies Jan 7 '17 at 17:05
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    $\begingroup$ The 4 and 8 are the same ones. For 2, there must be many choices because the curve is rational. $\endgroup$ – Noam D. Elkies Jan 7 '17 at 19:08
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    $\begingroup$ I've just checked on Wikipedia, they have several stunning identities involving fifth powers there for$$R(q)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}},$$e. g. $$R(q)^{-5}-R(q)^5=\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6$$or$$(v^4-3v^3+4v^2-2v+1)v=(v^4+2v^3+4v^2+3v+1)u^5$$with $u=R(q)$, $v=R(q^5)$, or$$(u^5+\phi^5)(v^5+\phi^5)=5\sqrt5\phi^5$$for $u=R(q^a)$, $v=R(q^b)$ with $5ab=4\pi^2$, or$$R(q)^5=w\left(\frac{1-w}{1+w}\right)^2,\ \ R(q^2)^5=w^2\frac{1+w}{1-w}$$with $w=R(q)R(q^2)^2$... $\endgroup$ – მამუკა ჯიბლაძე Jan 8 '17 at 8:44
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    $\begingroup$ @მამუკაჯიბლაძე: Yes, I've checked out the page as well. By the way, you made a small typo. It's $$R(q)^{-5}-R(q)^5 = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11$$ $\endgroup$ – Tito Piezas III Jan 8 '17 at 9:43
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    $\begingroup$ I think an answer to your question about quintic analogs of (1) and (3) is given in this paper arxiv.org/pdf/1304.0684.pdf $\endgroup$ – Nemo Jan 10 '17 at 17:30
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(Courtesy of a comment by Nemo who suggested Huber's paper.)

Part I. $x^5+y^5 = 1$

In "A Theory of Theta Functions to the Quintic base", Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ and the Rogers-Ramanujan continued fraction, $$R(z)=q^{1/5}\frac{H(q)}{G(q)}=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then Huber's four functions in simplified form are,

$$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ with golden ratio $\phi$. These obey, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ It then follows that the ratio of $a,b$ as well as $c,d$ are Rogers-Ramanujan cfracs, $$\frac{a(\tau)}{b(\tau)} = R(\tau)\tag3$$ $$\frac{c(\tau)}{d(\tau)} = \phi\, R\big(\tfrac{-1}{5\tau}\big)\tag4$$ so $R(\tau)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(5\tau)}{\phi+R(5\tau)}\tag5$$

Part II. $x^5+y^5+z^5 = 1$

It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. This is given by,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag6$$

where, $$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ and which can be derived by combining $(1),(2)$.

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    $\begingroup$ I agree with @Nemo that these $2/5$th powers spoil it a little bit, but it is awesome anyway! $\endgroup$ – მამუკა ჯიბლაძე Jan 12 '17 at 7:28
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    $\begingroup$ @მამუკაჯიბლაძე: The fractional powers disappear in the ratio $a/b$. I'm trying to check if $d/b$ can be simplified to a form so the eta gets "absorbed", but no luck so far. $\endgroup$ – Tito Piezas III Jan 15 '17 at 14:35
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    $\begingroup$ I see... this would then take care of $c/b=(c/d)(d/b)$ too $\endgroup$ – მამუკა ჯიბლაძე Jan 15 '17 at 14:39

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