Modular formulas in $X_0(2)$ Let $f_2(z) = \frac{\Delta(2z)}{\Delta(z)}$, where $\Delta$ is the modular discriminant and $z\in X_0(2)$. How can I prove that
$$\frac{(2E_2(2z) - E_2(z))^6}{\Delta(z)} = \frac{(1 + 2^6f_2(z))^3}{f_2(z)}\\
\frac{E_6^2(z)}{\Delta(z)} = \frac{(1 + 2^6f_2(z))(1 - 2^9f_2(z))^2}{f_2(z)}$$
? Thank you in advance
 A: I think you are missing some context here. The
Eisenstein series
are special modular forms for the modular group and of even weights.
The Eisenstein series $\,E_4,E_6\,$ are a basis for all the
other Eisenstein series except $\,E_2\,$ which is not a modular
form. However, $$E_{\gamma,2}(z) := 2E_2(2z) - E_2(z)$$ is a modular form.
Next, define the modular function
$$ f_2(z) := \Delta(2z)/\Delta(z) \quad \text{ where } \quad
  \Delta(z) := \eta(z)^{24}. $$
How to express $\,E_{\gamma,2}(z),E_4(z),E_6(z)\,$ in terms of
$\,\Delta(z),f_2(z)?\,$ One answer is
$$ E_{\gamma,2}(z)^6 = \Delta(z)\frac{(1 + 2^6f_2(z))^3}{f_2(z)}, \\
   E_4(z)^3 = \Delta(z)\frac{(1 + 2^8f_2(z))^3}{f_2(z)}, \\
   E_6(z)^2 = \Delta(z)\frac{(1 + 2^6f_2(z))(1 - 2^9f_2(z))^2}{f_2(z)}.
$$ How to prove these identities? One approach mentioned
in a comment is to use linear algebra on the appropriate
space of modular forms and compute a number of coefficients
of the $q$ expansions. This is very computational and does
depend on some knowledge about modular form spaces.
Another approach is to express everything in terms of the
Dedekind eta function. One way is
$$ E_{\gamma,2}(z) = \left(\frac{\eta(2z)^5}{\eta(z)\eta(4z)}
\right)^4 +2^4\left(\frac{\eta(4z)^2}{\eta(2z)}\right)^4, \\
   E_4(z) = \Big(1 + 2^8f_2(z)\Big)
 \left(\frac{\eta(z)^2}{\eta(2z)}\right)^8, \\
   E_6(z) = \Big(\eta(z)^8+2^5\eta(4z)^8\Big)\cdot \\ \frac{\Big(
\eta(z)^{16} -2^9\eta(z)^8\eta(4z)^8-2^{13}\eta(4z)^{16}
\Big)}{\eta(2z)^{12}}. $$
The expression for $\,E_4(z)\,$ already implies the identity
given earlier for $\,E_4(z).\,$ The expressions for $\,E_{\gamma,2}(z),E_6(z)\,$ combined with the well-known
identity $$ \eta(2z)^{24} = \eta(z)^{16}\eta(4z)^8 + 2^4\eta(z)^8\eta(4z)^{16} $$ imply the two other identities.
The problem now is how to prove these eta expressions are equal
to the corresponding Eisenstein series?
This depends very much on how the Eisenstein series are
defined and what is known about them. For example, the
$q$-expansion for $\,E_{\gamma,2}\,$ is
OEIS sequence A004011. In the
sequence entry is the comment

$E_{\gamma,2}$ is the unique normalized modular form for $\Gamma_0(2)$ of weight $2$.

If the eta expression can be shown to have the same property
then we are done. This could be applied to the other two cases
given knowledge about the unique properties of the Eisenstein
series.
