TITLE: Is $\eta^{24}(\tau)\,j(\tau) = {E_4}^3(q)$?
QUESTION [7 upvotes]: 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?:
$\begin{align}
1.\;\; \eta^{24}(\tau) &= \frac{{E_4}^3(q)}{j(\tau)}\\[1.5mm]
2.\;\; \eta^{24}(\tau) &= \frac{{E_6}^2(q)}{j(\tau)-1728}\\[1.5mm]
3.\;\; \eta^{48}(\tau) &= \frac{{E_8}^3(q)}{j^2(\tau)}\\
\end{align}$

REPLY [2 votes]: These relations are easily proved if we use the nome $q = e^{2\pi i\tau}$ and use Ramanujan notation of $P(q), Q(q), R(q)$ also.
We have
\begin{align}
\eta(\tau) &= \eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n})\tag{1}\\
E_{2}(\tau) &= P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}}\tag{2}\\
E_{4}(\tau) &= Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}\tag{3}\\
E_{6}(\tau) &= R(q) = 1 - 504\sum_{n = 1}^{\infty}\frac{n^{5}q^{n}}{1 - q^{n}}\tag{4}\\
E_{8}(\tau) &= Q^{2}(q) = 1 + 480\sum_{n = 1}^{\infty}\frac{n^{7}q^{n}}{1 - q^{n}}\tag{5}\\
j(\tau) &= \frac{1728E_{4}^{3}(\tau)}{E_{4}^{3}(\tau) - E_{6}^{2}(\tau)} = \frac{1728Q^{3}(q)}{Q^{3}(q) - R^{2}(q)}\tag{6}
\end{align}
Except for the fact $E_{8}(\tau) = Q^{2}(q) = E_{4}^{2}(\tau)$ mentioned in equation $(5)$ all the equations above are definitions of various functions of $q$ and corresponding functions of $\tau$.
It is easy to see that $$P(q) = 24q\frac{d}{dq}\{\log \eta(q)\}\tag{7}$$ And Ramanujan gives the fundamental differential equations
\begin{align}
q\frac{dP(q)}{dq} &= \frac{P^{2}(q) - Q(q)}{12}\tag{8}\\
q\frac{dQ(q)}{dq} &= \frac{P(q)Q(q) - R(q)}{3}\tag{9}\\
q\frac{dR(q)}{dq} &= \frac{P(q)R(q) - Q^{2}(q)}{2}\tag{10}
\end{align}
From these differential equations it is easy to see that
\begin{align}
q\frac{d}{dq}\{Q^{3}(q) - R^{2}(q)\} &= 3Q^{2}(q)q\frac{dQ(q)}{dq} - 2R(q)q\frac{dR(q)}{dq}\notag\\
&= 3Q^{2}(q)\frac{P(q)Q(q) - R(q)}{3} - 2R(q)\frac{P(q)R(q) - Q^{2}(q)}{2}\notag\\
&= P(q)Q^{3}(q) - R(q)Q^{2}(q) - P(q)R^{2}(q) + R(q)Q^{2}(q)\notag\\
&= P(q)\{Q^{3}(q) - R^{2}(q)\}\notag
\end{align}
and therefore $$q\frac{d}{dq}\{\log(Q^{3}(q) - R^{2}(q))\} = P(q)\tag{11}$$ Now comparing $(7)$ and $(11)$ we get $$Q^{3}(q) - R^{2}(q) = A\eta^{24}(q)$$ where $A$ is constant. Comparing coefficients of $q$ on both sides we can see that $$A = 3\cdot 240 + 2\cdot 504 = 1728$$ so that $$Q^{3}(q) - R^{2}(q) = 1728\eta^{24}(q)\tag{12}$$ Using this identity along with equation $(6)$ we get the first identity in question $$\eta^{24}(\tau) = \frac{E_{4}^{3}(\tau)}{j(\tau)}$$ and third identity in question is square of the first identity and the fact that $E_{8}(\tau) = E_{4}^{2}(\tau)$ as noted in equation $(5)$.
The second identity of question easily follows from equations $(6)$ and $(12)$.