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So it is well know that the Eisenstein series of weight 2 is not modular on $SL_2(\mathbb{Z})$.

In this paper of Kilford (http://uk.arxiv.org/PS_cache/math/pdf/0701/0701478v1.pdf) on page 4, he says that $E_2 \in M_2(\Gamma_0(2))$.

In fact we can obtain the following equality, $$ E_2 = \dfrac{\eta(2z)^{20}}{\eta(z)^8\eta(4z)^8} + 16\cdot \dfrac{\eta(4z)^8}{\eta(2z)^4} $$ (there's a typo in the orignal paper which is fixed here) and since the linear combination of eta-quotients on the right hand side is modular on $\Gamma_0(8)$ by a theorem of Ligozat (which can be found on page 2 of that same paper), we would have that $E_2$ is modular on $\Gamma_0(8)$ as well.

So my question is, do we have modularity for $E_2$ on $\Gamma_0(2)$ as well? I was wondering this since Kilford says that $E_2 \in M_2(\Gamma_0(2))$. If so, why? How about for $\Gamma_0(4)$?

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    $\begingroup$ Please note that if $f$ is a modular form for $\Gamma_0(n)$, then $f$ is a modular form for $\Gamma_0(kn)$ for all positive integers $k$. $\endgroup$
    – S. Carnahan ♦
    Commented Mar 27, 2012 at 8:17
  • $\begingroup$ Your $E_2$ is actually $E_{\gamma,2}$ whose $q$-series is the generating function of OEIS sequence A004011. $\endgroup$
    – Somos
    Commented Oct 21, 2017 at 23:52

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Kilford uses not $E_2$ but something he calls $E_{2, 2}$, which is $$E_{2}(z) - 2E_2(2z) = 1 - 24 \sum_{n \text{ odd}} \sigma_{1}(n) q^n.$$ This is a modular form of weight 2 and level $\Gamma_0(2)$. Similarly $E_{p, 2} = E_2(z) - p E_2(pz)$ is modular of level $\Gamma_0(p)$ for any $p$.

The naive Eisenstein series $$E_{2}(z) = 1 - 24 \sum_{n \in \mathbb{N}} \sigma_{1}(n) q^n$$ is not a modular form of any level.

[EDIT: My original answer contained the following statement, which is obviously wrong: "this follows from the fact that $E_2(-1/z) - z^2 E_2(z)$ is something like $6/\pi \mathrm{Im}(z)$, while for a modular form it would have to be holomorphic."

A hopefully better statement is: we have $E_2(z) - z^{-2} E_2(-1/z) = 2\pi i / z$, and if $E_2$ were modular (of some level) then $z^{-2} E_2(-1/z)$ would also be modular (of some other level) and hence their difference would be modular (for the intersection of the two level groups). But a non-constant rational function cannot be modular of any level.]

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  • $\begingroup$ How is this possible? $E_2(z)$ is holomorphic in $z$, isn't it. So also $z^2E_2(z)$ is holomorphic and also $E_2(-1/z)$ is also holomorphic (except at $z=0$). So then the difference has to be holomorphic as well. $\endgroup$
    – wood
    Commented Mar 27, 2012 at 10:48
  • $\begingroup$ In fact the formulas is $E_2(-1/z) - z^2 E_2(z)=-2 \pi i z$. But correcting with a non-holomorphic summand gives really modular transformation behavior. $\endgroup$
    – wood
    Commented Mar 27, 2012 at 10:58
  • $\begingroup$ Weight-two holomorphic Eisenstein series aren't defined directly by the usual series, which doesn't converge. Rather, consider a larger family of Eisenstein series with additional complex parameter $s$, show meromorphic continuation in $s$, and evaluate at $s_o$ most likely to produce something holomorphic in $z$. However, the obstruction is non-trivial in general. For Hilbert modular forms over fields other than $\mathbb Q$ the obstruction vanishes. This game is sometimes called "Hecke summation". The obstruction tends to be in the constant term, so the trick mentioned above succeeds. $\endgroup$
    – paul garrett
    Commented Mar 27, 2012 at 17:07
  • $\begingroup$ Right, if you do the Hecke trick, the non-holomorphic part will occur naturally. I am only saying that the difference of two holomorphic functions is holomorphic. In fact the correct transformation behavior $E_2(-1/z) - z^2 E_2(z)=-2 \pi i z$ also shows that $E_2$ cannot be modular for any level. $\endgroup$
    – wood
    Commented Mar 27, 2012 at 18:14
  • $\begingroup$ Sorry for the confusion, guys -- my original answer is obvious nonsense (I was somehow muddling the holomorphic and non-holomorphic $E_2$'s). I've replaced the bogus remark with an expansion of wood's last comment. $\endgroup$
    – David Loeffler
    Commented Mar 27, 2012 at 21:03

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