Direct proof of the level four eta identity

Question

Of Somos' over 6200 eta identities, the one of smallest level (maybe the simplest?) is the eta identity of level $4$ $$\eta(z)^{16} \eta(4z)^8 + 16 \eta(4z)^{16} \eta(z)^8 = \eta(2z)^{24}.$$

As with all identities among modular forms, we can prove it by checking the first few coefficients. I wonder if it is possible to give a "real" proof of this straight from the product $$\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1 - q^n)$$ in the style of this answer. The higher exponents in this problem seem to make it harder to use Jacobi's triple product.

Answer 1

Consider the product representations of the Jacobi Thetanulls (as functions of the period ratio $\tau$): $$\begin{align} \varTheta_{00}(\tau) &= \frac{\eta(\tau)^5}{\eta\left(\frac{\tau}{2}\right)^2\eta(2\tau)^2} & \varTheta_{01}(\tau) &= \frac{\eta\left(\frac{\tau}{2}\right)^2}{\eta(\tau)} & \varTheta_{10}(\tau) &= \frac{2\eta(2\tau)^2}{\eta(\tau)} \end{align}$$ and plug these into Jacobi's Vierergleichung $$\varTheta_{00}^4 = \varTheta_{01}^4 + \varTheta_{10}^4$$ This results in $$\frac{\eta(\tau)^{20}}{\eta\left(\frac{\tau}{2}\right)^8\eta(2\tau)^8} = \frac{\eta\left(\frac{\tau}{2}\right)^8}{\eta(\tau)^4} + \frac{16\,\eta(2\tau)^8}{\eta(\tau)^4}$$ Multiplying out the denominators and setting $\tau=2z$ then results in the identity you asked about.

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One way of proving the Vierergleichung works by first deducing the period-doubling/-halving relations $$\begin{align} \varTheta_{00}(\tau)^2 &= \varTheta_{00}(2\tau)^2+\varTheta_{10}(2\tau)^2 & \varTheta_{00}(2\tau)^2 &= \frac{1}{2} \left(\varTheta_{00}(\tau)^2+\varTheta_{01}(\tau)^2\right) \\ \varTheta_{01}(\tau)^2 &= \varTheta_{00}(2\tau)^2-\varTheta_{10}(2\tau)^2 & \varTheta_{01}(2\tau)^2 &= \varTheta_{00}(\tau)\,\varTheta_{01}(\tau) \\ \varTheta_{10}(\tau)^2 &= 2\,\varTheta_{00}(2\tau)\,\varTheta_{10}(2\tau) & \varTheta_{10}(2\tau)^2 &= \frac{1}{2} \left(\varTheta_{00}(\tau)^2-\varTheta_{01}(\tau)^2\right) \end{align}$$ The middle left equation has been deduced in my answer to the question you referenced. The upper left then follows from replacing $\tau$ by $\tau+1$. Addition/subtraction reveals the upper and lower right identities. The middle right identity easily follows from its left neighbor once you plug in $$\begin{align} \varTheta_{00}(\tau) &= \varTheta_{00}(4\tau)+\varTheta_{10}(4\tau) \\ \varTheta_{01}(\tau) &= \varTheta_{00}(4\tau)-\varTheta_{10}(4\tau) \end{align}$$ which is just a splitting of $q$-series into odd and even terms. Finally, plugging in the above two even/odd-splitting equations in the lower right identity yields the lower left identity.

Anyway, once you have established the above set of identities, you can deduce $$\begin{align} \varTheta_{10}(\tau)^4 &= 4\,\varTheta_{00}(2\tau)^2\,\varTheta_{10}(2\tau)^2 \\ &= \left(\varTheta_{00}(\tau)^2+\varTheta_{01}(\tau)^2\right) \left(\varTheta_{00}(\tau)^2-\varTheta_{01}(\tau)^2\right) \\ &= \varTheta_{00}(\tau)^4-\varTheta_{01}(\tau)^4 \end{align}$$ which is the Vierergleichung.

Answer 2

I am surprised at the idea that this needs "real" proof. The identity is the very first one in my list and the very first formulation of it is the "aequatio identica satis abstrusa" straight from Jacobi's Fundamenta Nova Theoriae Functionum Ellipticarum, page 90 equation (14) published in 1829. There is a reference to Bruce Berndt's editing of Ramanujan's Notebooks, Part III, page 40, Entry 25 (vii) which is exactly the same that is used in the parametrization of the Arithmetico-Geometric Mean by squares of the theta null functions. That is, $\;\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)\;$ is the $a^2=b^2+c^2$ of the AGM and this goes back to Gauss even before Jacobi.

Answer 3

This identity

$$\eta(z)^{16} \eta(4z)^8 + 16 \eta(4z)^{16} \eta(z)^8 = \eta(2z)^{24}$$

written as

$$\bigg(\frac{\eta(z)}{\eta(2z)}\bigg)^{16}\bigg(\frac{\eta(4z)}{\eta(2z)}\bigg)^8+16\bigg(\frac{\eta(z)}{\eta(2z)}\bigg)^8\bigg(\frac{\eta(4z)}{\eta(2z)}\bigg)^{16}=1$$

is equal respectively, to

$$\beta_{4n}+\alpha_{4n}=1.$$

Using Ramanujan's notations for

$\eta(z)=\frac{\phi(q)}{\sqrt{2}(G_{n})^2}$

$\eta(2z)=\frac{\phi(q^2)}{\sqrt{2}(G_{4n})^2}$

$\eta(4z)=\frac{\phi(q^4)}{\sqrt{2}(G_{16n})^2}$

and that

$\frac{\phi(q)}{\phi(q^2)}=\frac{\beta_{4n}^{1/4}}{\beta_{n}^{1/8}}$

$\frac{\phi(q^2)}{\phi(q^4)}=\frac{\beta_{16n}^{1/4}}{\beta_{4n}^{1/8}}$

and moreover

$\beta_{n}=\frac{g_{n}^8}{G_{n}^8}$

$\beta_{4n}=\frac{g_{4n}^8}{G_{4n}^8}$

$\beta_{16n}=\frac{g_{16n}^8}{G_{16n}^8}$

$$4g_{4n}^8 G_{4n}^8\big(G_{4n}^8-g_{4n}^8\big)=1\tag1$$

we arrive to

$\beta_{4n}=\bigg(\frac{\eta(z)}{\eta(2z)}\bigg)^{16}\bigg(\frac{\eta(4z)}{\eta(2z)}\bigg)^8=\frac{\beta_{4n}^{5}G_{4n}^{48}}{\beta_{n}^2 G_{n}^{32} \beta_{16n}^2 G_{16n}^{16}}=$

$=\frac{g_{4n}^{40} G_{4n}^8}{g{n}^{16} G_{n}^{16} g_{16n}^{16}}=$

$=\frac{g_{4n}^{40} G_{4n}^8}{g_{n}^{16} G_{n}^{16} 2^4 g_{4n}^{16} G_{4n}^{16}}=$

$=\frac{g_{4n}^{40} G_{4n}^8}{g_{4n}^{16} g_{4n}^{16} G_{4n}^{16}}$

simplifying

$\beta_{4n}=\frac{g_{4n}^8}{G_{4n}^8}$

With analog considerations we have

$$\frac{g_{4n}^8}{ G_{4n}^8}+\frac{1}{4g_{4n}^8 G_{4n}^{16}}=1$$

or that is the same the equation (1).