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Given the Dedekind eta function,

$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$

where $q = \exp(2\pi i\tau)$. Consider the following "family",

$\begin{align} \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{24} &= \frac{u^8}{(-1+16u^8)^2},\;\;\; u = q^{1/8} \prod_{n=1}^\infty \frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})^2}\\[2.5mm] \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{12} &= \frac{c^3}{(1+c^3)(-1+8c^3)^2},\;\;c = q^{1/3} \prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\\[2.5mm] \left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}\; &= \frac{r^5}{(r^5+u_5^5)(r^5-u_5^{-5})},\quad r\; =\; q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\[2.5mm] \left(\frac{\eta(7\tau)}{\eta(\tau)}\right)^{4}\;&= \frac{h(h-1)}{1+5h-8h^2+h^3},\quad h = 1/q\, \prod_{n=1}^\infty \frac{(1-q^{7n-2})^2(1-q^{7n-5})^2(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-1})^3(1-q^{7n-6})^3}\\[2.5mm] \left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} &=\frac{s}{(s-u_{13})(s+u_{13}^{-1})},\quad\; s =\; ???\\ \end{align}$

with fundamental units $u_n$ as $u_5 = \frac{1+\sqrt{5}}{2}$ and $u_{13} = \frac{3+\sqrt{13}}{2}$. The second-to-the-last appears in Chap 10 (10.2) of Duke's Continued Fractions and Modular Functions.

Question: What is the analogous infinite product, if any, for $\left(\frac{\eta\,(13\tau)}{\eta\,(\tau)}\right)^2$ similar to the ones above?

Postscript: This question has been modified before, as it was a bit unclear.

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  • $\begingroup$ Dear Tito, What exactly do you want to know? The ratio $\eta(13\tau)^2/\eta(\tau)^2$ is a modular function on $\Gamma_0(13)$, and, since it is a ratio of infinite products, it is expressible as an infinite product. Since $\Gamma_1(13)$ has genus $2$ (not $0$, unlike in the cases for $2,3,5$ and $7$ that you write down in your question), it probably can't be written in a non-trivial way as a rational function of a modular function on $\Gamma_1(13)$. Regards, $\endgroup$ – Matt E Jul 1 '12 at 15:18
  • $\begingroup$ @MattE: I should not have mentioned $\Gamma_1(13)$ since I was not sure of its genus. Anyway, if we disregard that aspect of the question, what I really want to know is this: What is the EXPLICIT expression of the eta quotient $Q_{13} = (\eta(13\tau)/\eta(\tau))^2$ in terms of a polynomial in $P_{13}$, where $P_{13}$ is an infinite product similar to the ones above? I want to complete the "family" above, if possible. $\endgroup$ – Tito Piezas III Jul 1 '12 at 16:18
  • $\begingroup$ Dear Tito, What is wrong with the obvious formula $(\eta(13\tau)/\eta(\tau)\bigr)^2 = q \prod_{n =1}^{\infty} (1+ q^n + \cdots + q^{12n})^2$, which follows directly from the product formula for $\eta$? Regards, $\endgroup$ – Matt E Jul 1 '12 at 22:52
  • $\begingroup$ @MattE: I knew somebody was eventually going to bring that up. It is too obvious. I am hoping for a product that shows a somewhat more "surprising" relationship, and akin to the ones above. $\endgroup$ – Tito Piezas III Jul 2 '12 at 5:56
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This answer collects my comments above, and relates them to David Loeffler's answer:

In the examples given, you are writing down a uniformizer for the genus $0$ modular curve $X_0(N)$ (with $N = 2, 3, 5$, and $7$), as well as a uniformizer for some genus $0$ cover of $X_0(N)$, and the algebraic relationship between the two uniformizers on the two different modular curves.

Now the function $\bigl(\eta(13\tau)/\eta(\tau)\bigr)^2$ is a uniformizer on $X_0(13)$ (which is again a curve of genus $0$). However, since $X_1(13)$ already has genus two (unlike e.g. $X_1(N)$ for $N = 2, 3, 5,$ and $7$, each of which have genus $0$), I don't see any obvious modular curve which is a proper cover of $X_0(13)$ but which still has genus $0$, and hence which allows one to generalize the other examples.

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  • $\begingroup$ Thanks, Matt. I chose yours since you did answer first, as well as pointed out that while $X_0 (N)$ for N = {2,3,5,7, 13} has the same genus, $X_1(N)$ for N = 13 has a different genus from the others. It is reassuring to know that while those 5 primes indeed have something in common which explains certain eta quotients, they also have differences which explains the absence of N = 13 in the family above. $\endgroup$ – Tito Piezas III Jul 2 '12 at 13:47
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Michael Somos just today found the identity,

$$ \left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$

where,

$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(1-q^{13n-12})} $$

thus completing the family for $N = 2,3,5,7,13$.

Does this address Matt E and Loeffler's comments? Is "s" a modular function?

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In all your identities, the left-hand side is the standard(-ish) modular function of level $X_0(p)$ giving an isomorphism of $X_0(p)$ onto $\mathbf{P}^1$ (for the five primes $p$ where such a thing exists). And the other side corresponds to some "nice" uniformizer of a genus 0 modular curve that is a finite cover of $X_0(p)$; maybe there is some specific pattern to how you're choosing these but it's not so clear to me. So your question amounts, as far as I can tell, to:

Does there exist a modular function that generates a nontrivial finite extension of the function field of $X_0(13)$ and has a "nice" infinite product formula?

That's kind of a hard question to answer without a precise definition of "nice".

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  • $\begingroup$ Well, it is hard to define "nice", but I believe all nice formulas are in The Book :-) $\endgroup$ – Tito Piezas III Jul 2 '12 at 14:22

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