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I am new to number theory, so I am guessing this is a standard formula. I would be grateful for a reference:

We know that the Eisenstein series $G_2$ is quasimodular of level $SL_2(\mathbb Z)$, so the derivative $G'_2$ is also quasimodular. As such, we can express $G'_2$ as a polynomial in $G_2$ with modular coefficients (of level $SL_2(\mathbb Z)$. Could anyone please tell me the coefficients?

In other words: we know

$$G'_2=a_0+a_1G_2+a_2G_2^2$$

where $a_0$, $a_1$ and $a_2$ are modular forms of level $\Gamma(1)=SL_2(\mathbb Z)$. Can you please tell me the expressions of $a_0$, $a_1$ and $a_2$?

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According to Ramanujan (and Wikipedia) $qdE_2/dq = (E_2^2-E_4)/12$, where $q=e^{2\pi i\tau}$. It should convert to what you want.

http://en.wikipedia.org/wiki/Eisenstein_series

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It's not too bad to check these out by hand, at least for a few low-degree examples such as $E_2$.

To be fair, we should be clear about normalizations before we begin: I will use the normalizations $$ E_2 = 1 - 24q - 72q^2 - 96q^3 + \cdots $$ and $$ E_4 = 1 + 240q + 2160q^2 + 6720q^3 + \cdots $$ although you can pick your favourite normalization. In such a case we find that $$ DE_2 = q\frac{d}{dq}E_2 = -24q - 144q^2 - 288q^2 + \cdots $$ and that $$ E_2^2 = 1 - 48q + 432q^2 + \cdots $$ Since we know that $E_2^2$ and $E_4$ will generate the degree 4 terms, we just need to check the first two coefficients. That is, we know that $DE_2 = \alpha E_2^2 + \beta E_4 = (\alpha + \beta) + (-48\alpha + 240\beta) + \cdots$ which then becomes a straightforward linear algebra problem: that is, we are trying to solve $$ \alpha + \beta = 0 \\ -48\alpha + 240\beta = -24 $$ i.e. $\beta = -\frac{1}{12}$ and $\alpha = \frac{1}{12}$.

I suppose then to fully answer your question: We would pick $a_0 = -\frac{1}{2}E_4$, $a_1 = 0$, and $a_2 = \frac{1}{12}$.

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  • $\begingroup$ Of course, different normalizations will yield different coefficients... $\endgroup$
    – Simon Rose
    Commented Aug 6, 2014 at 13:17
  • $\begingroup$ Is it clear apriori that $DE_2$ is a modular form of weight $4$? After all, $E_2$ is only quasimodular. $\endgroup$
    – GH from MO
    Commented Aug 6, 2014 at 15:06
  • $\begingroup$ Well, $DE_2$ is necessarily quasimodular (not modular) of pure weight 4. It isn't too hard to check that differentiation a quasimodular form of pure weight $k$ will yield a quasimodular form of pure weight $k + 2$ (essentially, see what happens when you differentiate the expression $f(\gamma \tau) = (c\tau + d)^k f(\tau)$). $\endgroup$
    – Simon Rose
    Commented Aug 7, 2014 at 13:42
  • $\begingroup$ Although, if you wanted to verify this, one thing you could do is do the same computation but using the first three (or perhaps 4, if you want to be super careful) coefficients instead of only the first two. If you consider the space of quasi-modular forms of weight less than 4, then this is 3 or 4 dimensional (depending on if you include the weight 0 part). So you need three/four generators, which are given by $E_2, E_2^2, E_4$ (and perhaps the constant term). This is a longer computation, but if you do it, you'll see that you get the same answer. $\endgroup$
    – Simon Rose
    Commented Aug 7, 2014 at 13:44
  • $\begingroup$ Thank you! Is there a good reference about quasi-modular forms? $\endgroup$
    – GH from MO
    Commented Aug 7, 2014 at 14:52
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Section 1.3 of Zagier's notes express $G_2'$, $G_4'$, $G_6'$ as explicit elements of $\mathbb{C}[G_2,G_4,G_6]$. In particular, (15) shows that your equation holds with $$ a_0=\frac{5\pi i}{3}G_4,\qquad a_1=0,\qquad a_2=-4\pi i.$$

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