9
$\begingroup$

Let $q = e^{2\pi i \tau}$. Given the j-function,

$j = j(q) = 1/q + 744 + 196884q + 21493760q^2 + \dots$

and define,

$k = j-1728$

Let $\tau =\sqrt{-N}$, where $N > 1$. Anybody knows how to prove the RHS of these conjectured relations?:

$\begin{align}q^{-1/60} G(q) = q^{-1/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})} &= j\,^{1/60}\,_2F_1\left(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\right)\\ &= k\,^{1/60}\,_2F_1\left(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{-1728}{k}\right)\\[2.5mm] q^{11/60} H(q) = q^{11/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})} &= j\,^{-11/60}\,_2F_1\left(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\right)\\ &= k\,^{-11/60}\,_2F_1\left(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{-1728}{k}\right)\end{align}$

  • 1
    I've edited the post to include $G(q)$ and $H(q)$.2014-01-08

1 Answers 1

4

Let \begin{align} g(\tau) &= q^{-1/60} G(q) \\ h(\tau) &= q^{11/60} H(q) \end{align}

First of all, the equalities between the hypergeometric $j$ and $k$ expressions follow from the eulerian transformation ${}_2F_1(a,b;c;z) = (1-z)^{-b}\,{}_2F_1\left(c-a,b;c;\frac{z}{z-1}\right)$ Therefore it suffices to prove the identities between $g$ resp. $h$ and the corresponding hypergeometric $j$ expressions. I will translate those to more familiar identities.

We will use the Rogers-Ramanujan continued fraction (RRCF), $\rho(\tau) = \frac{h(\tau)}{g(\tau)} = q^{1/5}\frac{H(q)}{G(q)}$ Formula $(22)$ from the above MathWorld entry on RRCF essentially states that $\frac{1}{\rho^{5}} - 11 - \rho^5 = \frac{1}{g^6\,h^6}$ (Use the product representation of $g$ and $h$ to identify the right-hand side). From this, we can easily deduce \begin{align} g &= \frac{1} {\left(\rho - 11\,\rho^6 - \rho^{11}\right)^{1/12}} \\ h &= \frac{\rho} {\left(\rho - 11\,\rho^6 - \rho^{11}\right)^{1/12}} \end{align} assuming that the arguments to the radicals are small positive reals, which should follow from the restrictions you have placed on $\tau$.

Furthermore, formula $(46)$ from the above Mathworld entry on RRCF gives the relation of $\rho$ with Klein's $j$: $j = \frac {\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^3} {\left(\rho - 11\,\rho^6 - \rho^{11}\right)^5}$ which allows us to write \begin{align} g\,j^{-1/60} &= \left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^{-1/20} \\ h\,j^{11/60} &= \frac {\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^{11/20}} {1 - 11\,\rho^5 - \rho^{10}} \end{align} again assuming that the arguments to the radicals are small positive reals.

I need a deus ex machina now, and Raimundas Vidūnas arXiv:0807.4808v1 comes to the rescue. His formulae $(59)$ and $(61)$ in section 6.3 ("icosahedral hypergeometric equations", p. 20) state precisely that \begin{align} {}_2F_1\left(\tfrac{19}{60},-\tfrac{1}{60};\tfrac{4}{5};\varphi_1(x)\right) &= \left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^{-1/20} \\{}_2F_1\left(\tfrac{31}{60},\tfrac{11}{60};\tfrac{6}{5};\varphi_1(x)\right) &= \frac{\left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^{11/20}} {1 + 11\,x - x^2} \\\text{where}\qquad \varphi_1(x) &= \frac{-1728\,x\,(1+11\,x-x^2)^5} {\left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^3} \end{align} And your conjecture follows from setting $x=-\rho^5$ which implies $\varphi_1(x) = \frac{1728}{j}$.

Summarizing, there is an algebraic relation between $j$ and $g$ resp. $h$, and the hypergeometric ${}_2F_1$ expressions are designed to solve those algebraic relations for $g\,j^{-1/60}$ resp. $h\,j^{11/60}$, given $j$.

  • 0
    You suspect correctly. :) However, I've partially forgotten some of the reasoning I used in this post, so it will take me some time to retrace my steps.2014-02-10