Given $\large q=e^{2\pi i \tau}$. Define, $$\alpha(\tau) = \sqrt2\,q^{1/8}\prod_{n=1}^\infty\frac{ (1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$ $$\beta(\tau) = q^{1/5}\prod_{n=1}^\infty\frac{ (1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$ $$\gamma(\tau) = q^{1/4}\prod_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-2})(1-q^{6n-4})}$$ $$\delta(\tau) = q^{1/3}\prod_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}$$ $$\epsilon(\tau) = q^{1/2}\prod_{n=1}^\infty\frac{ (1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}$$ $$\lambda(\tau) = q^{1/1}\prod_{n=1}^\infty\frac{ (1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}$$

each of which has a beautiful $q$-continued fraction. Then we propose the functional equations,

$$\begin{aligned} \alpha^8\Big(\frac{-1}{4\tau}\Big)&=1-\alpha^8(\tau)\\[2mm] \beta^5\Big(\frac{-1}{5\tau}\Big)&=\frac{\phi^5\beta^5(\tau)-1}{-\beta^5(\tau)-\phi^5},\quad\phi=\tfrac{1+\sqrt5}2\\[2mm] \gamma^4\Big(\frac{-1}{6\tau}\Big)&=\frac{\gamma^4(\tau)-\tfrac19}{\gamma^4(\tau)-1}\\[2mm] \delta^3\Big(\frac{-1}{6\tau}\Big)&=\frac{\delta^3(\tau)-\tfrac18}{-\delta^3(\tau)-1}\\[2mm] \epsilon^2\Big(\frac{-1}{8\tau}\Big)&=\frac{u^2\,\epsilon^2(\tau)-1}{\epsilon^2(\tau)-u^2},\quad u =1+\sqrt2\\[2mm] \lambda\Big(\frac{-1}{12\tau}\Big)&=\frac{v\,\lambda(\tau)-1}{\lambda(\tau)-v},\quad v = 2+\sqrt3\end{aligned}$$

Note 1: Excepting $\gamma^4\big(\tfrac{-1}{6\tau}\big)$ which is in this post and served as the model, I just found the rest empirically, but their consistent forms suggest these are correct.

Note 2: Also, the two order $6$ obey the simple $\displaystyle \frac1{\gamma^4(\tau)}-\frac1{\delta^3(\tau)}=1$.


  1. How do we derive the easiest one from first principles and generalize it for the rest?
  2. Is the list complete? Without requiring representation as a q-continued fraction, is there a similar functional equation for, say, a $p=7$ order?
  • $\begingroup$ The transformation for $\beta^5$ is equation (7.3) p. 150 of W. Duke, Continued fractions and modular functions, Bull. Amer. Math. Soc. 42 (2005) 137-162. In pag. 154 of the same paper it is found the one for $\epsilon^2$. $\endgroup$
    – juan
    Feb 4, 2017 at 19:26
  • 1
    $\begingroup$ @juan: Thanks. I have read that paper many years ago and didn't realize it contained two of the relations above. $\endgroup$ Feb 5, 2017 at 1:20

1 Answer 1


Throughout this answer, I'll be referencing Kubert-Lang, "Modular Units" Chapter 2, sections 1 and 2, and Chapter 3 section 4.

Each of your functions is a ratio of the Siegel functions of the form $$A_N(\tau)=\left(\frac{g_{\frac1N,0}}{g_{\frac aN,0}}(N\tau)\right)^m.$$

If $r=(r_1,r_2)\in \mathbb Q^2$, $B_2(x)=x^2-x+1/6$ is the second Bernoulli polynomial, $q=e^{2\pi i \tau}$ as usual, and $\zeta=e^{2\pi i r_2}$, then the Siegel function $g_a$ is defined by $$ g_a(\tau)= -q^{\frac12 B_2(r_1)}e^{2\pi i \frac{r_2(r_1-1)}2}\prod_{n=0}^\infty(1-q^{n+r_1}\zeta)(1-q^{n+1-r_1}\zeta^{-1}). $$ The Siegel functions $g_a$ are modular units, meaning they're modular functions whose zeros and poles are supported at cusps.

If $\gamma\in \text{SL}_2(\mathbb Z)$, then $$ g_a|_0\gamma=g_{a\gamma}. $$ Moreover, we can reduce $r_1$ and $r_2$ modulo $\mathbb Z$, or change signs of both $r_1$ and $r_2$, if we introduce a root of unity. However, the $m$'s in your ratios satisfy $m(1-a^2)\equiv 0 \pmod N$ if $N$ is odd and $\pmod{2N}$ if $N$, and so using Theorem 3.4.1 of Kubert-Lang we find we can ignore this extra factor so the transformations rules becomes this simple. Note, the theorem is written in terms of Klein functions $t_a=g_a\Delta^{-\frac1{12}}$. Using these transformation rules, it's easy to see that all of your functions are invariant under $\Gamma_1(N)/\{\pm1\}$.

The specific groups you have considered are ones where this group is genus $0$, and your functions $A_N$ are Hauptmoduln with a unique zero at the cusp $\infty$, a constant $C_0$ at the cusp $0$, and a unique pole at one other cusp. We can calculate $C_0$ using the formulas above: If $\zeta_{2N}=e^{2\pi i \frac{1}{2N}}$, then the constant is given by $\left(\frac{\zeta_{2N}-\zeta_{2N}^{-1}}{\zeta_{2N}^{a}-\zeta_{2N}^{-a}}\right)^m$.

Since $A_N$ is a hauptmodl, its image under the Fricke involution is a rational function in $A_N$. Since it now has a unique $0$ at the cusp $0$ (Where $A_N$ has the constant $C_0$) and the constant $C_0$ at the cusp infinity (Where $A_N$ has a zero), we must have $$ A_N|_0\begin{pmatrix}0&-1\\N&0\end{pmatrix}=\frac{-A_N+C_0}{X\cdot A_N+1}, $$ for some $X$, determined by the movement of the pole. If the pole does not move, then $X=0$. Otherwise, $-1/X$ is the constant of $A_N$ at the cusp where $A_N|_0\begin{pmatrix}0&-1\\N&0\end{pmatrix}$ has its pole.

All in all, this suggest you might have hope for $N=7,9$ and $10$, which all give genus $0$ groups. Unfortunately The ratios of the type you've described above aren't hauptmoduln. You might be able to find similar results if you allow for larger order products.


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