# Is $\eta^{24}(\tau)\,j(\tau) = {E_4}^3(q)$?

Given the j-function $j(\tau)$,

$j(\tau) = 1728J(\tau)$,

where $J(\tau)$ is Klein’s absolute invariant, the Dedekind eta function $\eta(\tau)$, and the following Eisenstein series,

\begin{align} E_4 (q) &= 1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n}\\[1.5mm] E_6 (q) &= 1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\\[1.5mm] E_8 (q) &= 1+480\sum_{n=1}^{\infty} \frac{n^7q^n}{1-q^n}\\ \end{align}

where,

$q = \exp(2\pi i \tau)$

Are the following relations true?:

\begin{align} 1.\;\; \eta^{24}(\tau) &= \frac{{E_4}^3(q)}{j(\tau)}\\[1.5mm] 2.\;\; \eta^{24}(\tau) &= \frac{{E_6}^2(q)}{j(\tau)-1728}\\[1.5mm] 3.\;\; \eta^{48}(\tau) &= \frac{{E_8}^3(q)}{j^2(\tau)}\\ \end{align}

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These are not too difficult to prove. See my answer. –  glebovg Sep 6 at 8:33

Yes, these three relations are all true. Since $\eta^{24}$ is the weight 12 level 1 cusp form $\Delta$, you can write them as relations between level 1 modular forms, and these are easy to check because the relevant modular form spaces have small finite dimensions.

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Thank you, David. –  Tito Piezas III Mar 20 '12 at 15:29

The first and third identity are easy to prove. Recall that the modular discriminant $\Delta = \Delta(\omega_1, \omega_2)$ is defined by $$\Delta = g_2^3 - 27g_3^2,$$ where $g_2 = g_2(\omega_1, \omega_2) = 60G_4(\omega_1, \omega_2)$ and $g_3 = g_3(\omega_1, \omega_2) = 140G_6(\omega_1, \omega_2)$ are the Weierstrassian invariants. Moreover, we define the Klein invariant $J = J(\omega_1, \omega_2)$ and the $j$-function by $$J = \frac{g_2^3}{\Delta} \quad \text{and} \quad j = 12^3 J,$$ respectively.

Let $\lambda \ne 0$. Because $g_2$ and $g_3$ are of order $-4$ and $-6$, respectively, we have $$g_2(\lambda \omega_1, \lambda \omega_2) = \lambda^{-4} g_2 \quad \text{and} \quad g_3(\lambda \omega_1, \lambda \omega_2) = \lambda^{-6} g_3,$$ and $$\Delta(\lambda \omega_1, \lambda \omega_2) = \lambda^{-12} \Delta,$$ accordingly. Moreover, since $g_2^3$ and $\Delta$ are of the same order, $J(\lambda \omega_1, \lambda \omega_2) = J(\omega_1, \omega_2)$. In particular, $$g_2(\tau) = \omega_1^4 g_2, \quad g_3(\tau) = \omega_1^6 g_3, \quad \text{and} \quad \Delta(\tau) = \omega_1^{12} \Delta$$ On the other hand, $$J = J(\tau) \quad \text{and} \quad j = j(\tau),$$ that is to say, both $J$ and $j$ are functions of $\tau$ alone.

Using the classical identity $\Delta(\tau) = (2\pi)^{12} \eta^{24}(\tau)$, where $\eta(\tau)$ is the Dedekind eta function defined by $$\eta(\tau) = q^{1/12} \prod_{n = 1}^\infty (1 - q^{2n}), \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ we obtain $$\eta^{24}(\tau) = \frac{\Delta(\tau)}{(2\pi)^{12}} = \frac{\omega_1^{12} g_2^3}{(2\pi)^{12} J(\tau)} = \frac{12^3 \omega_1^{12} g_2^3}{(2\pi)^{12} j(\tau)}.$$ In view of a known result, $$E_{2k}(q) = 1 - \frac{4k}{B_{2k}} \sum_{n = 1}^\infty \frac{n^{2k - 1} q^{2n}}{1 - q^{2n}}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ where $B_{2k}$ are the Bernoulli numbers and $E_{2k}(q) = G_{2k}(\tau)/2\zeta(2k)$, we get $$g_2(\tau) = 60G_4(\tau) = \frac{4\pi^4 E_4(q)}{3},$$ or, what is the same thing, $$g_2 = \omega_1^{-4} \frac{4\pi^4 E_4(q)}{3},$$ Therefore, $$\eta^{24}(\tau) = \frac{12^3 \omega_1^{12}}{(2\pi)^{12} j(\tau)} \left(\omega_1^{-4} \frac{4\pi^4 E_4(q)}{3}\right)^3 = \frac{E_4^3(q)}{j(\tau)}.$$ Lastly, using the fact that $E_8(q) = E_4^2(q)$ leads to $$\eta^{48}(\tau) = \frac{E_8^3(q)}{j^2(\tau)}.$$

Now, the second identity follows from the first and the fact that $$J(\tau) = \frac{E_4^3(q)}{E_4^3(q) - E_6^2(q)}.$$ Indeed, we have $$E_6^2(q) = E_4^3(q) - 12^3 \frac{E_4^3(q)}{j(\tau)},$$ but in view of the first identity this becomes $$E_6^2(q) = (j(\tau) - 12^3)\eta(\tau)^{24},$$ and the result follows.

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