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The weight-2 Eisenstein series $$E_{2}(z) := 1 - 24\sum_{n\geq 1} \sigma_{1}(n)q^{n}$$ is not a modular form, but is "quasi-modular" in the sense that $E_{2}(-1/z)$ is equal to $z^{2}E(z) +$(error term). It is known that for $f$ a modular form of weight $k$ that $q\frac{d}{dq} f$ is a modular form modulo $p$ for chosen prime $p$. Indeed, consider $R := \left(q\frac{d}{dq}f - \frac{k}{12}E_{2}f\right)E_{p-1} + \frac{k}{12}E_{p+1}f$ and the relations $E_{p-1} \equiv 1 \pmod{p}$ and $E_{p+1} \equiv E_{2} \pmod{p}$.

My question is: is $q\frac{d}{dq}E_{2}$ a modular form modulo $p$ for some prime $p$? Or is it maybe some kind of automorphic form in some other way?

reuns
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Freddie
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  • Please clarify the transformation formula for $E_2$ and that $q\frac{d}{dq}f - \frac{k}{12}E_{2}f$ is a modular form (of what weight?) – reuns Apr 13 '23 at 10:49
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    The answer by @reuns is 100% correct, but perhaps a little disappointing. Perhaps a little more surprising is that $q \tfrac{\mathrm{d}}{\mathrm{d}q} E_2$ is a mod $p^n$ modular form for every $n$; equivalently, it is a p-adic modular form (in the sense of Serre). – David Loeffler Apr 13 '23 at 23:24
  • @DavidLoeffler Oh this is interesting. Do you have any references that I could check out to read more about this? – Freddie Apr 14 '23 at 16:48

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Isn't $E_{2,p}(z)=E_2(z)-p E_2(pz)$ a (level $p$) modular form for any $p$?

So apply your formula to $f=E_{2,p} \equiv E_2 \bmod p$

reuns
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