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How can we prove that $$\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4) \ ?$$

Here, $\eta(q)$ is the Dedekind Eta Function, which is defined by

$$\eta(q)=q^{1/24}\prod_{n=1}^\infty \left(1-q^{2n} \right) $$

I looked up some identities of Dedekind Eta Function in the hope of simplifying the right hand expression but I failed. I would be grateful if any one could suggest a proof of this identity.


I came across this identity in a paper. They refer to Entry $t_{8,12,48}$ and $t_{8,18,60a}$ of Dedekind eta function product identities. Unfortunately, I was unable to find these entries on that site.

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Where did you come across this problem? –  Steven Stadnicki Sep 4 '13 at 5:42
$\eta^n(x)$ means the same as $\eta(x)^n$? –  Dan Brumleve Sep 4 '13 at 5:49
@DanBrumleve: Yes, that is what I meant. –  Shobhit Sep 4 '13 at 5:54
Shobhit: To find these entries, 1) go to the main page that you linked; 2) choose the 4th sub-link that says "All eta function product identities by level"; 3) then the 3rd link "level 8 has 116 identities"; 4) and the identity you want is the second one, labeled as t8_12_48. –  Tito Piezas III Sep 5 '13 at 4:25

1 Answer 1

up vote 4 down vote accepted

Usually, the Dedekind eta function is understood as a function of $\tau\in\mathbb{H}$ (complex upper half plane) where $q=\mathrm{e}^{2\pi\mathrm{i}\tau}$. This way, the modular symmetries of $\eta$ can be expressed easily and the branching issues with $q^{1/24}$ can be avoided. In this answer, I will not make use of modular symmetries, but perhaps some other answer will, so let us stick to the convention of regarding $\eta$ as a function of $\tau$ and use another symbol when we regard it as a function of $q$. Following Michael Somos's example at the product identity website linked in the question, I will write $h(q)$ for $\eta(\tau)$.

Dividing by $h^4(q^4)\,h^4(q^2)$ and isolating the rightmost summand, your identity in question becomes $$\frac{h^{10}(q^4)}{h^4(q^2)\,h^4(q^8)} - \frac{4\,h^4(q^8)}{h^2(q^4)} = \frac{h^4(q)}{h^2(q^2)} \tag{*}$$

Apart from having some power of $q$ as overall multiplier, the factors of eta products (and quotients) can be written as $1+a_n$ with $a_n=\operatorname{O}(q^{kn})$ for some $k>0$ with $k$ and the implied $\operatorname{O}$ constant not depending on $n$, therefore $\sum_{n=1}^\infty |a_n|$ converges absolutely for $|q|<1$, and we can reorder the factors accordingly. Using rules like $$\begin{aligned} \prod_{n=1}^\infty (1-x^{2n}w)(1-x^{2n-1}w) &= \prod_{n=1}^\infty (1-x^n w) \\ \prod_{n=1}^\infty (1+x^n)(1-x^{2n-1}) &= 1 \end{aligned}$$ for $x=q^k$ and applying Jacobi's triple product identity, we find $$\begin{aligned} \frac{h^{10}(q^4)}{h^4(q^2)\,h^4(q^8)} &= \prod_{m=1}^\infty (1-q^{4m})^2\,(1+q^{4m-2})^4 = \left(\sum_{n\in\mathbb{Z}} q^{2n^2}\right)^2 = \vartheta_3^2(q^2) \\ \frac{4\,h^4(q^8)}{h^2(q^4)} &= q\prod_{m=1}^\infty (1-q^{4m})^2\,(1+q^{4m})^2\,(1+q^{4m-4})^2 = q\left(\sum_{n\in\mathbb{Z}} q^{2n(n+1)}\right)^2 = \vartheta_2^2(q^2) \\ \frac{h^4(q)}{h^2(q^2)} &= \prod_{m=1}^\infty (1-q^{2m})^2\,(1-q^{2m-1})^4 = \left(\sum_{n\in\mathbb{Z}} (-q)^{n^2}\right)^2 = \vartheta_4^2(q) \end{aligned}$$ where $\vartheta_2, \vartheta_3, \vartheta_4$ are Jacobi thetanull functions. Therefore (*) is equivalent to the identity $$\vartheta_3^2(q^2) - \vartheta_2^2(q^2) = \vartheta_4^2(q) \tag{**}$$ This is one of six formulae that relate theta nullvalues with those corresponding to halved or doubled period ratios. It can be shown by reordering the nested sum for $\vartheta_4^2(q)$ in a chessboard-like manner: $$\begin{aligned} \vartheta_4^2(q) &= \sum_{u,v\in\mathbb{Z}} (-q)^{u^2+v^2} = \sum_{\substack{r,s\in\mathbb{Z}\\r+s\text{ even}}} q^{r^2+s^2} - \sum_{\substack{r,t\in\mathbb{Z}\\r+t\text{ odd}}} q^{r^2+t^2} \\ &\stackrel{\substack{r = k - m\\s = k + m\\t = k + m + 1}}{=} \sum_{k,m\in\mathbb{Z}} q^{2(k^2+m^2)} - \sum_{k,m\in\mathbb{Z}} q^{\left((2k+1)^2+(2m+1)^2\right)/2} = \vartheta_3^2(q^2) - \vartheta_2^2(q^2) \end{aligned}$$ You may be interested in writing down the other period-ratio doubling and halving formulae and transforming them to eta product identities that involve power products of $h(q), h(q^2), h(q^4), h(q^8)$ only.

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Thank you very much! –  Shobhit Sep 8 '13 at 7:55
Does this imply that $\phi (\sqrt 2 q)- q \psi (q^2) = $\phi (-q)$ where the $\psi(q)$ and \phi(q)$ are Ramanujan theta functions ? –  Shivam Patel Dec 24 '13 at 7:32
@ShivamPatel: I can only guess what you mean. There are easy odd/even decompositions of theta series like $\vartheta_3(q) = \vartheta_3(q^4) + \vartheta_2(q^4)$ which are basic power series manipulations. Then there is Jacobi's Vierergleichung $\vartheta_3^4(q) = \vartheta_2^4(q)+\vartheta_4^4(q)$ which can indeed be deduced from the period-doubling/-halving formulae. –  ccorn Dec 24 '13 at 10:44

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