On the Klein quartic and the similar $a^2b+b^2c+c^2a$?

Question

Given the Ramanujan theta function,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$

Let $q = e^{2\pi i \tau}$ and assume $\tau = \sqrt{-d}$.

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I. Degree 5

\begin{align} a &= q^{11/60}\;\frac{f(-q,-q^4)}{f(-q)} \,=\, q^{11/60}\, \prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}\\ b &= q^{-1/60}\,\frac{f(-q^2,-q^3)}{f(-q)} = q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}\\ \end{align}

where $a,b$ surprisingly are radicals. They obey the nice relation,

$$a\,b^{11}-a^{11}\,b = 11a^6\,b^6 +1$$

Equivalently,

$$\Big(\frac{b}{a}\Big)^5-\Big(\frac{a}{b}\Big)^5 = \frac1{(ab)^6}+11$$

The integer $60$ reflects the order of the icosahedron.

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II. Degree 7

\begin{align} a &= -q^{61/168}\;\frac{f(-q,-q^6)}{f(-q^2)} = -q^{61/168}\prod_{n=1}^\infty \frac{(1-q^{7n-1})(1-q^{7n-6})(1-q^{7n})}{(1-q^{2n})}\\ b &= \,q^{13/168}\;\frac{f(-q^2,-q^5)}{f(-q^2)} \,=\; q^{13/168}\;\prod_{n=1}^\infty\frac{(1-q^{7n-2})(1-q^{7n-5})(1-q^{7n})}{(1-q^{2n})}\\ c &= q^{-11/168}\;\frac{f(-q^3,-q^4)}{f(-q^2)} = q^{-11/168}\,\prod_{n=1}^\infty \frac{(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n})}{(1-q^{2n})}\\ \end{align}

where $a,b,c$ again are radicals. They obey,

$$a^3b+b^3c+c^3a = 0$$

and the integer $168$ reflects the order of the Klein quartic. (See also this post.)

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III. Degree 9

\begin{align} a &= \;q^{5/9}\;\frac{f(-q,-q^8)}{f(-q^3)} = q^{5/9}\prod_{n=1}^\infty \frac{(1-q^{9n-1})(1-q^{9n-8})(1-q^{9n})}{(1-q^{3n})}\\ b &= \;q^{2/9}\;\frac{f(-q^2,-q^7)}{f(-q^3)} \,=\, q^{2/9}\prod_{n=1}^\infty \frac{(1-q^{9n-2})(1-q^{9n-7})(1-q^{9n})}{(1-q^{3n})}\\ c &= -q^{-1/9}\;\frac{f(-q^4,-q^5)}{f(-q^3)} = -q^{-1/9}\prod_{n=1}^\infty \frac{(1-q^{9n-4})(1-q^{9n-5})(1-q^{9n})}{(1-q^{3n})} \end{align}

where $a,b,c$ still are radicals. They now obey,

$$\color{blue}{a^2b+b^2c+c^2a = 0}$$ $$a^2c+b^2a+c^2b = 1$$

(Added later.) The first is equivalent to,

$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c} = 0$$

so the ratio of the proper functions sum to zero which does not happen with the lower degrees. More on "Nonic Analogues of the Rogers-Ramanujan functions" (though they didn't affix the factor $q^{m/9}$ to make them into radicals).

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IV. Degree 11

Studied by G. Andrews in 1975, the first of five has the form,

$$a_1=\frac{f(-q,-q^{10})}{f(-q^4)}$$

To make it a radical, I affixed a factor $q^{m/n}.\,$ I hoped it would involve $n = 660$ thus manifesting V. Arnold's trinities regarding PSL(2,5), PSL(2,7), and PSL(2,11) with orders $60, 168, 660$, but it was only $n = 264 = 11\times24$. (Sigh.)

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V. Questions

1. Is there anything else special about $a^2b+b^2c+c^2a = 0$?
2. In what other contexts does it appear?

Answer 1

The questions are:

1. Is there anything else special about $a^2b+b^2c+c^2a = 0$?
2. In what other contexts does it appear?

For question 1, Given the $q$-series of degree $9$, $$ \frac ac=a^3-b^3,\quad\frac ba=b^3-c^3,\quad\frac cb=c^3-a^3 $$ which leads to the telescoping sum $$ \frac ac+\frac ba+\frac cb = 0 $$ as noticed already. Another feature is multi-section. Let $$ A := -q^{-1/9}\frac{f(-q^{1/3})}{f(-q^3)} = a + b + c $$ where $a,b,c$ form a 3-section of the $q$-series $A$.

For question 2, refer to L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXI, section "Impossibility of $x^3+y^3=z^3$", pp. $545$-$550$. On page $546$ it states

$\quad$ J. A. Euler$^9$ noted that, if $\,p^3+q^3+r^3=0\,$ is possible, $\,x=p^2q,$ $y=q^2r,$ $z=r^2p\,$ satisfy $\,x/y+y/z+z/x=0\,$ or $\,x^2z+y^2x+z^2y=0.$

Something similar happens with the Klein quartic curve and the Fermat septic curve.

Answer 2

From the geometrical point of view, there is something special about the equation $$ F = a^2b + b^2c + c^2a - \lambda abc $$ for a varying parameter $\lambda$ (for which you have $\lambda=0$). Namely this equation defines a rational surface $X = \{F=0\}\subset \mathbb{P}^2_{a,b,c}\times \mathbb{P}^1_\lambda$ and, after resolving singularities, this surface admits an extremal semistable elliptic fibration $f\colon X \to \mathbb{P}^1_\lambda$ or, in other words, it is a rational elliptic fibration with only $I_n$ fibres and the fewest possible number of singular fibres (which are of type $I_9$, $I_1$, $I_1$, $I_1$ in this case).

The classification of extremal semistable rational elliptic fibrations is due to Beauville [1], and this example appears as the last entry in his table. In particular he obtains it as a quotient $X = \mathbb{C}\times\mathfrak{h} \, / \, \Gamma$ using the following subgroup of $\operatorname{SL}(2,\mathbb{Z})$ $$ \Gamma = \Gamma_0(9)\cap\Gamma_0^0(3)=\left\{ \begin{pmatrix} a & b \\ c& d \end{pmatrix} \in \operatorname{SL}(2,\mathbb{Z}) : \begin{matrix} a\equiv 1\mod 3 \\ c\equiv 0\mod 9 \end{matrix}\right\},$$ so there should be some way to make a link with modular forms via this description. (But I am not the person to explain it.)

[1] A. Beauville, Les familles stables de courbes elliptiques sur $\mathbb{P}^1$ admettant quarte fibres singulières. C. R. Acad. Sci. Paris 294 (1982), 657–660.

Answer 3

Thanks to those who answered. Doing more research, while browsing the book "Generic Polynomials" (thanks, Rouse!), in page 30 I saw the generic cubic for $C_3 = A_3$,

$$x^3 + m x^2 - (m + 3)x + 1 = 0$$

which has negative square discriminant $D = -(m^2+3m+9)^2,$ hence all roots $x_i$ are real. After some experimentation, if we define,

$$a = x_1^{1/3},\quad b = x_2^{1/3},\quad c = x_3^{1/3}$$

then we have,

$$\color{blue}{a^2b+b^2c+c^2a = 0}$$ $$a^2c+b^2a+c^2b \neq 0$$

so the correct order of roots $x_i$ of the generic cubic must be chosen. IF TRUE, this explains the properties below for levels $9, 13, 27$.

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I. Level 9

The level $9$ theta quotients satisfy the cubic,

$$x^3+(u+3)x^2-(u+6)x+1=0$$

which, after minor change of variables, is just the generic cubic in disguise. With $u = \left(\frac{\eta(\tau)}{\eta(9\tau)}\right)^3$, the roots are,

$$x_1 = \frac{\eta(3\tau)}{\eta(\tau)} \left(\frac{\eta(3\tau)}{\eta(9\tau)}\right)^3 \left(\frac{q^{5/9}\,f(-q,-q^8)}{\quad f(-q^3)}\right)^3$$

$$x_2 = \frac{\eta(3\tau)}{\eta(\tau)} \left(\frac{\eta(3\tau)}{\eta(9\tau)}\right)^3 \left(\frac{q^{2/9}\,f(-q^2,-q^7)}{\quad f(-q^3)}\right)^3$$

$$x_3 = \frac{\eta(3\tau)}{\eta(\tau)} \left(\frac{\eta(3\tau)}{\eta(9\tau)}\right)^3 \left(\frac{-\,f(-q^4,-q^5)}{q^{1/9}\,f(-q^3)\;}\right)^3$$

with $f(a,b)$ the usual Ramanujan theta function. Taking their cube roots $\sqrt[3]{x_i}$, they then obey the equation in blue, a relation mentioned in the original post above.

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II. Level 13

As discovered by Ramanujan, the level $13$ theta quotients satisfy the cubic,

$$x^3+(v+1)x^2-(v+4)x+1=0$$

which, after minor change of variables, is also the generic cubic in disguise. With $v = \left(\frac{\eta(\tau)}{\eta(13\tau)}\right)^2$, the roots are,

$$x_1 = \left(\frac{f(-q^2,-q^{11})}{f(-q,-q^{12})}\right) \left(\frac{f(-q^{10},-q^{3})}{f(-q^5,-q^{8})}\right)$$

$$x_2 = \frac{-1}{\,q}\left(\frac{f(-q^4,-q^{9})}{f(-q^2,-q^{11})}\right) \left(\frac{f(-q^{6},-q^{7})}{f(-q^3,-q^{10})}\right)$$

$$x_3 = \frac{1}{q^{-1}}\left(\frac{f(-q^8,-q^{5})}{f(-q^4,-q^{9})}\right) \left(\frac{f(-q^{12},-q)}{f(-q^6,-q^{7})}\right)$$

Taking their cube roots $\sqrt[3]{x_i}$, they obey the equation in blue.

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III. Level 27

The Dyson Mod 27 Identities also apply, at least three of them. (Caveat: I changed the order in the Mathworld link.) Those satisfy the cubic,

$$x^3+(w+3)x^2-(w+6)x+1=0$$

which is just a variation for level 9 but with $w = \left(\frac{\eta(3\tau)}{\eta(27\tau)}\right)^3$. The roots are,

$$x_1 = \frac{\eta(9\tau)}{\eta(3\tau)} \left(\frac{\eta(\tau)}{\eta(27\tau)}\right)^3 \left(\frac{q^{2}\,f(-q^3,-q^{24})}{\quad f(-q)}\right)^3$$

$$x_2 = \frac{\eta(9\tau)}{\eta(3\tau)} \left(\frac{\eta(\tau)}{\eta(27\tau)}\right)^3 \left(\frac{q\,f(-q^6,-q^{21})}{\quad f(-q)}\right)^3$$

$$x_3 = \frac{\eta(9\tau)}{\eta(3\tau)} \left(\frac{\eta(\tau)}{\eta(27\tau)}\right)^3 \left(\frac{-\,f(-q^{12},-q^{15})}{f(-q)\;}\right)^3$$

Taking their cube roots $\sqrt[3]{x_i}$, they also obey the equation in blue. (I've ignored the fourth Dyson Mod 27 Identity, so a more satisfying relation is a quartic with coefficients determined by some eta quotient.)

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IV. Conclusion

Just like the Klein quartic $a^3b+b^3c+c^3a$, it seems there is more to,

$$a^2b + b^2c+ c^2a = 0$$

than meets the eye. Does it have a name, so it is Internet searchable?