Relations of eisenstein series with eta quotient

Question

Theorem 1.67 On page 19 of Ken Ono's book The Web of Modularity says:

Every modular form on $SL_2(\mathbb{Z})$ may be expressed as a rational function in $\eta(z)$, $\eta(2z)$ and $\eta(4z)$.

The proof uses the following two facts,

1, As graded algebra, $M(SL_2(\mathbb{Z}))$ is generated by $E_4(z)$ and $E_6(z)$.

2, $E_4(z)=\frac{n(z)^{16}}{\eta(2z)^8}+2^8\frac{\eta(2z)^{16}}{\eta(z)^8}$,

$E_6(z)=\frac{\eta(z)^{24}}{\eta(2z)^{12}}-2^5\cdot3\cdot5\cdot\eta(2z)^{12}-2^9\cdot3\cdot11\cdot\frac{\eta(2z)^{12}\eta(4z)^8}{\eta(z)^8}+2^{13}\frac{\eta(4z)^{24}}{\eta(2z)^{12}}$.

My question is how did people discover the two identities?

Answer 1

This does not answer your question, but I want to elaborate on my comment to Jeremy's answer that such identities can also be understood from the viewpoint of elliptic functions. I use as a reference Vol. 2 of Erdelyi et al., Higher Transcendental functions. By Eq. (13.20.8), $$g_2=C\left(\theta_2^8+\theta_3^8+\theta_4^8\right), $$ where $\theta_k=\theta_k(0|\tau)$ are Jacobi theta values, $C$ is irrelevant and $g_2$ is essentially your $E_4$. There is a similar formula for $g_3\sim E_6$. The theta values can be expressed in terms of Dedekind's eta function (this has a priori nothing to do with modular forms, it's just a matter of rewriting the classical product formulas in new notation). With $z=\tau/2$ we have $$\theta_2=\frac{2\eta(4z)^2}{\eta(2z)},\quad \theta_3=\frac{\eta(2z)^5}{\eta(z)^2\eta(4z)^2},\quad \theta_4=\frac{\eta(z)^2}{\eta(2z)}. $$ (I took this from Wikipedia, but it can be found in many books.) This already gives a formula expressing $E_4$ in terms of $\eta(z)$, $\eta(2z)$ and $\eta(4z)$, but not quite the one you ask about.

To get closer, we recall Jacobi's quartic identity (13.14.23), $$\theta_2(0|\tau)^4+\theta_4(0|\tau)^4=\theta_3(0|\tau)^4$$ and (13.23.15), which tells us that $$\theta_2(0|2\tau)^2=\frac 12\left(\theta_3(0|\tau)^2-\theta_4(0|\tau)^2\right), $$ $$\theta_3(0|2\tau)^2=\frac 12\left(\theta_3(0|\tau)^2+\theta_4(0|\tau)^2\right). $$ Multiplying together and using the quartic identity gives $$ \theta_2(0|2\tau)^2\theta_3(0|2\tau)^2=\frac 14\left(\theta_3(0|\tau)^4-\theta_4(0|\tau)^4\right)=\frac 14\theta_2(0|\tau)^4.$$ We then obtain from the quartic identity $$\theta_2(0|\tau)^8+\theta_3(0|\tau)^8=\theta_4(0|\tau)^8+2\theta_2(0|\tau)^4\theta_3(0|\tau)^4= \theta_4(0|\tau)^8+\frac 18\,\theta_2(0|\tau/2)^8.$$ Thus, we get the alternative formula $$g_2=C\left(2\theta_4(0|\tau)^8+\frac 18\,\theta_2(0|\tau/2)^8\right). $$ This seems to give $$E_4=\frac{\eta(z)^{16}}{\eta(2z)^8}+2^4\frac{\eta(2z)^{16}}{\eta(z)^8}. $$ Note that Ono has $2^8$ instead of $2^4$. I don't know if the reason is a trivial mistake in my calculations, in the references I used or in Ono's book.

My conclusion is that even though Ono's formulas may be new, the problem he solves is very classical and was first treated using the theory of elliptic and theta functions rather than the much newer theory of modular forms.

Answer 2

The answer is probably some fairly basic linear algebra. For the first one, each term on the right hand side is a modular form of weight $4$ and level $2$. The space $M_{4}(\Gamma_{0}(2))$ has dimension $2$ and it is a finite computation to enumerate the eta quotients in this space. There are two, and they span the space.

There are no eta-quotients of level $2$ and weight $6$ (because $\Gamma_{0}(2)$ has an elliptic point of order $2$), so it is natural to look for weight $6$ forms of level $4$. (It is a classical result that every level four modular form can be expressed in terms of eta quotients - see Theorem 1.49 in Ono's book.) The space $M_{6}(\Gamma_{0}(4))$ has dimension $4$ and contains $10$ eta quotients. Pick your favorite four that span the space, and express $E_{6}(z)$ in terms of them.