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For even $k \geq 4$, the Eisenstein series \begin{align*} G_k(\tau) &= \sum_{(n, m)\in \mathbb{Z}^2} \frac{1}{(m + n\tau)^k} \end{align*} (omitting the term $(n, m) = (0, 0)$) is a modular form of weight $k$ for $\Gamma = SL_2(\mathbb{Z})$, and in fact the ring of modular forms of $\Gamma$ is just $\mathbb{C}[E_4, E_6]$. The usual proof involving rearranging the double series over $n$ and $m$, which fails to converge compactly for the case $k = 2$. Still, $G_2$ is reasonably close to a modular form; it satisfies $G_2(-1/\tau) = \tau^2 G_2(\tau) + \alpha \tau$ for some constant $\alpha\not = 0$, and the graded ring $M_* = \mathbb{C}[G_2, G_4, G_6]$ is closed under the operator $D \vert M_k = \frac{1}{2\pi i} \frac{d}{dq} + \beta k$ for some constant $\beta$.

All this is easy enough to prove directly, but is there some deeper reason why $G_2$ is a quasimodular form? That is, why is $G_2$ still very close to being a modular form despite the bad behavior of the series above for $k = 2$; or, conversely, why doesn't $\Gamma$ have any modular forms of weight $2$ despite having a reasonable candidate in $G_2$?

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The only modular form for ${\rm SL}_2(\mathbf Z)$ of weight $2$ is $0$ because the orbit space for ${\rm SL}_2(\mathbf Z)$ acting on the upper half-plane is a sphere with one missing point and the sphere has genus $0$.

For a congruence subgroup $\Gamma$ of ${\rm SL}_2(\mathbf Z)$ there is a notion of weight $k$ modular forms for this group, and the space of them is denoted $M_k(\Gamma)$. Without getting into the details, this is a finite-dimensional complex vector space and it has a dimension formula that you can find in Theorem 36 here. Taking $k = 2$, it turns out that $\dim(M_2(\Gamma)) = g-1+C$, where $g$ is the genus of the modular curve $X(\Gamma)$ and $C$ is the number of cusps of $X(\Gamma)$. When $\Gamma = {\rm SL}_2(\mathbf Z)$, $X(\Gamma)$ is a sphere, which has genus $g = 0$, and there is one cusp, so $C = 1$. Thus $M_2({\rm SL}_2(\mathbf Z))$ has dimension $0-1+1 = 0$, so $M_2({\rm SL}_2(\mathbf Z)) = \{0\}$.

For prime $p$, $X_0(p)$ has $2$ cusps, so $\dim(M_2(\Gamma_0(p))) = g-1+C = g+1$, so $M_2(\Gamma_0(p))$ is not $\{0\}$. As pointed out on the MSE page here, $E_2(\tau)- pE_2(p\tau) \in M_2(\Gamma_0(p))$ with constant term $(p-1)/24$ in its $q$-expansion, so that's a nonzero example of weight $2$ for $\Gamma_0(p)$. For $p = 2, 3, 5$, and $7$, $X_0(p)$ has genus $0$, so $M_2(\Gamma_0(p))$ consists of scalar multiples of $E_2(\tau) - pE_2(p\tau)$. The curve $X_0(11)$ has genus $1$, so $M_2(\Gamma_0(11))$ has dimension $g+1=2$. A cuspidal weight $2$ modular form for $\Gamma_0(11)$ is on the same MSE page mentioned above.

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  • $\begingroup$ Thanks, that makes sense. I get that Riemann-Roch should put a bound on $M_k(\Gamma)$, but is there any inherent reason why I should expect $G_2$ to be quasimodular rather than just some random function? Or is the idea simply that the proof of the modularity of $G_k$ almost goes through for $k = 2$, and quasimodularity is just what you get from the parts of the proof that are still salvageable in that case? $\endgroup$
    – anomaly
    Commented Apr 20, 2023 at 14:10
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    $\begingroup$ Since the proof you cite for the quasi-modularity relation between $G_2(\tau)$ and $G_2(-1/\tau)$ involves swapping the order of a conditionally convergent double sum, you might like to see an alternate proof presented as a sequence of exercises on the page ctnt-summer.math.uconn.edu/wp-content/uploads/sites/1632/2016/…. $\endgroup$
    – KCd
    Commented Apr 22, 2023 at 0:53
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    $\begingroup$ Concerning a conceptual role for $G_2$, I can't offer one, but Kilford's book on modular forms mentions in Section 2.8.1 work of Kaneko and Zagier on an extension of modular forms that includes $G_2$: M. Kaneko and D. Zagier, A generalized Jacobi theta function and quasi-modular forms, pp. 165-172 in "The moduli space of curves" (Texel Island, 1994), Birkhauser Boston, 1995. $\endgroup$
    – KCd
    Commented Apr 22, 2023 at 0:55
  • $\begingroup$ Very interesting, thanks. I'll take a look at the alternative proof and track down a copy of that book. $\endgroup$
    – anomaly
    Commented Apr 22, 2023 at 1:28

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