Are there any obvious reasons $g_2^3-27g_3^2$ a cusp form? Recall that $G_k(\tau)$ are Eisenstein series$(k\geq 2)$ defined over upper half plane $\mathcal{H}$. Now define $g_2=60G_4,g_3=140G_6$ and $\Delta=(g_2)^3-27(g_3)^2$. Note that $\Delta$ corresponds to elliptic curve's discriminant defined by Wierstrass form $\mathcal{P}'^2=\mathcal{P}^3-g_2\mathcal{P}-g_3$.
$\textbf{Q:}$ Are there obvious reasons that $\Delta$ is a cusp form? I can compute fourier expansion constant term vanishing but this relies heavily upon $\zeta(2k)$'s concrete form in terms of Bernouille numbers. 
Ref. Diamond, Schurman A First Course in Modular Forms(4th Ed), Ex 1.7(d).

There is a complex torus/elliptic curve reason.


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*$z \mapsto (\wp_\tau(z),\wp_\tau'(z))$ is an isomorphism $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z}) \to E_\tau/\mathbb{C} : y^2 = 4x^3-g_2(\tau) x-g_3(\tau)$ (with $20 g_2(\tau),28 g_3(\tau)$ the coefficients of $z^2, z^4$ in the Laurent expansion of $\wp_\tau(z)$ at $z=0$)

*The defining series of $g_2(\tau),g_3(\tau)$ are holomorphic on the upper-half-plane,  $1$-periodic and weight $4,6$ invariant under $z \mapsto -1/z$, and since they converge as $\tau \to i\infty$ (absolute convergence of the defining series) they are modular forms  $\in M_4(SL_2(\mathbb{Z})),M_6(SL_2(\mathbb{Z}))$ and $\Delta(\tau) = g_2(\tau)^3-27g_3(\tau)^2 \in M_{12}(SL_2(\mathbb{Z}))$.

*$\Delta(\tau) = \prod_{j<l} (e_j(\tau)-e_l(\tau))$ is the discriminant of $4x^3-g_2(\tau) x-g_3(\tau) = 4 \prod_{j=1}^3(x-e_j(\tau))$ so $\Delta(\tau) = 0 $ implies the cubic  has a double root and $E_\tau$ is not an elliptic curve. Thus $\Delta(\tau)$ is non-zero on the upper-half plane.

*Therefore $j(\tau)= j(E_\tau)= 1728\frac{g_2(\tau)^3}{\Delta(\tau)}$ is  meromorphic on the modular curve and non-constant and holomorphic on the upper-half plane. Whence it has a pole at the unique cusp of the modular curve that is at $i \infty$. 

*And hence that $g_2(i\infty)= 60 \sum_{m \ne 0} \frac{1}{m^4} \ne 0$  implies that $\Delta(i\infty)= 0$ ie. $\Delta$ is a cusp form.
