# Why is the Eisenstein series $G_2$ a quasimodular form?

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 $$G_2$$?

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.

• 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? Commented Apr 20, 2023 at 14:10
• 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/….
– KCd
Commented Apr 22, 2023 at 0:53
• 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.
– KCd
Commented Apr 22, 2023 at 0:55
• Very interesting, thanks. I'll take a look at the alternative proof and track down a copy of that book. Commented Apr 22, 2023 at 1:28