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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?:

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

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How are the hypergeometric functions connected with $G(q)$ and $H(q)$ –  Shivam Patel Jan 8 at 9:23
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I've edited the post to include $G(q)$ and $H(q)$. –  Tito Piezas III Jan 8 at 16:08
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1 Answer

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 $j$.

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Tito Piezas III: I might have stumbled upon the Vidūnas preprint by following one of your links, and the MathWorld entries referenced here contain stuff contributed by you. Therefore I entertain the suspicion that you are familiar with all of the above and rather want to see one or more of the lemmas proven upon which this answer hinges. If so, please comment. :-) –  ccorn Feb 10 at 21:16
    
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. –  Tito Piezas III Feb 10 at 21:21
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