On V. Arnold's trinities regarding PSL(2,5), PSL(2,7), and PSL(2,11)?

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 $\tau = \sqrt{-d}.$ Then the following functions for levels $p=5, 7, 11, 13$ are radicals,

$$q^{11/60} \frac{f(-q,-q^4)}{f(-q)}$$ $$q^{61/168} \frac{f(-q,-q^6)}{f(-q^2)}$$ $$q^{199/264} \frac{f(-q,-q^{10})}{f(-q^4)}$$ $$q^{149/156} \frac{f(-q,-q^{12})}{f(-q^5)}$$

(Note: The numerators for levels $p=11,13$ are corrected courtesy of Somos.) Without the $q$-factor, they are analogues of the Rogers-Ramanujan identities, the first being the namesake, and which are sum-products,

$$\sum_{n=0}F_1(q) = \prod_{n=1}F_2(q)$$

Three of the integers involved, namely $60, 168, 156,$ are orders of transitive groups which in Magma notation are $5T4, 7T5, 13T6$. The one for $p=11$ is the odd one out since $264$ is not a group order. But we can use alternative quotients, such as the one for level $7$,

$$q^{17/42} \frac{f(-q,-q^6)}{\color{blue}{f(-q)}}$$

which is also a radical and $42$ is the order of $7T4$ (and the answer to the Hitchhiker's universe). It is not necessary that there be a sum-product associated with it.

Question: For level $p=11$, can we find an alternative quotient (both numerator and denominator) such that, together with the particular factor $q^{m/660},$ then it is also a radical?

P.S. The desired integer $660$ is the order of $11T5$ so this would be another manifestation of V. Arnold's trinities regarding PSL(2,5), PSL(2,7), and PSL(2,11). I've tried various denominators $f(-q^n)$ but can't get $q^{m/660}$.

Answer 1

This is a tentative answer requested by Tito.

Starting in 2008 I worked on a collection of Ramanujan theta function identities as an extension of my Dedekind eta product identities collection. In order to do this effectively, I needed to know what power of $q$ factor to use which "completes" these functions. My source was Bruce Berndt, Ramanujan Notebooks, Part III, page $42$ which states

$ G(q) = q^{(m-n)^2/8(m+n)}f(q^m,q^n) .$

For example, if $q=e^{2\pi i\tau}$, then $\eta(k\tau) = q^{k/24}f(-q^k) = q^{k/24}f(-q^k,-q^{2k}),$ and the power of $q$ factor is thus $q^{k/24}$.

For another example, $$h_1 = \frac{q^{121/104}f(-q,-q^{12})}{q^{5/24}f(-q^5)} = q^{149/156}\frac{f(-q,-q^{12})}{f(-q^5)} = \\ q^{149/156}(1 - q + q^5 + 2q^{10} - 2q^{11} - q^{12} + 4q^{15} +\dots) .$$

Another source for $p=13$ is R. J. Evans, Theta Functions Identities, $1990$, pages $97, 99, 113, 116$.

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(By TP). As allowed by the OP, the formula for the exponent $d$ of the $q$-factor

$$q^d\, f(q^m,q^n)$$

for level $13$ seems to be,

$$d(m,n) = \frac{(m-n)^2}{8(m+n)}-5/24$$

This yields,

$$d(1,12) = \frac{149}{156}, \quad d(2,11) = \frac{89}{156}, \quad d(3,10) = \frac{41}{156}$$ $$\quad d(4,9) = \frac{5}{156},\quad d(5,8) = -\frac{19}{156}, \quad d(6,7) = -\frac{31}{156}$$

This leads to the ratio,

$$r_2 = \frac{h_4}{h_2} = \frac{q^{5/156}f(-q^4,-q^{9})}{q^{89/156}f(-q^2,-q^{11})} = q^{-7/13}\frac{f(-q^4,-q^{9})}{f(-q^2,-q^{11})}$$

and the reduced $q$ factor matches the one in Ramanujan's ratio formula. This implies the $q$ factors in the addendum are in error because the wrong pairings were used.

Answer 2

(This is not really an answer, but addresses the question by Somos. Caveat: It turns out his version of the six $q$ factors in the other answer are the correct ones.)

To find the powers of the $q$ factor, my method uses two parts: one is to use a formula by Ramanujan and second is to find an appropriate eta quotient. For example, for $p=13$, apparently,

$$h_1 = q^{179/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}, \quad\quad h_2 = q^{119/156}\;\frac{f(-q^2,-q^{11})}{f(-q^5)}$$ $$h_3 = q^{35/156}\;\frac{f(-q^3,-q^{10})}{f(-q^5)},\quad\quad h_4 = q^{-37/156}\;\frac{f(-q^4,-q^{9})}{f(-q^5)}$$ $$h_5 = q^{-61/156}\;\frac{f(-q^5,-q^{8})}{f(-q^5)},\quad\quad h_6 = q^{-1/156}\;\frac{f(-q^6,-q^{7})}{f(-q^5)}$$

Their product is,

$$\prod_{k_1}^6 h_i = \frac{\eta(\tau)}{\eta(5\tau)}\left(\frac{\eta(13\tau)}{\eta(5\tau)}\right)^5$$

while their ratios have neat $q$-factor powers,

$$r_1 = \frac{h_2}{h_1} = q^{-5/13}\,F_1,\quad\quad r_2 = \frac{h_3}{h_2} = q^{-7/13}\,F_2$$ $$r_3 = \frac{h_4}{h_3} = q^{-6/13}\,F_3,\quad\quad r_4 = \frac{h_5}{h_4} = q^{-2/13}\,F_4$$ $$r_5 = \frac{h_6}{h_5} = q^{5/13}\,F_5,\quad\quad r_6 = \frac{h_1}{h_6} = q^{15/13}\,F_6$$

consistent with Ramanujan's ratio formula (disregarding signs) for $p=13$,

$$r_k = (-1)^{k-1}q^{k(3k-p)/(2p)}\,\frac{f(-q^{2k},-q^{p-2k})}{f(-q^{k},-q^{p-k})}$$

Edit:

It turns out that since levels $p=11, 13$ involves more functions than $p=7$, I inadvertently used the wrong pairs to derive the $q$ factor. Thus, the correct version is by Somos,

$$\text{Somos} = q^{149/156}\;\frac{f(-q,-q^{12})}{f(-q^5)}$$

The product given above remains accurate though.