TITLE: Serre's formula for $\Delta^{1/3}$
QUESTION [8 upvotes]: Let
$$E\quad\colon \quad y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$
be an elliptic curve, and let $b_4 = a_1a_3+2a_4$.
Serre in Propriétés galoisiennes des points d'ordre fini des courbes elliptiques  (Section 5.3, b), p. 305) gives the following formula
$$\Delta^{1/3}=b_4-3(x_i x_j+x_kx_l),\label{1}\tag{1}$$
where the $x_i$ are the $x$-coordinates of points of $E$ of order $3$, and $\{1,2,3,4\}=\{i,j\}\cup\{k,l\}$. This formula can be rewritten as
\begin{equation*}
 j^{-1/3}=\frac{\frac{1}{2}\frac{27j}{j-12^3}+3(x_ix_j+x_kx_l)}{\frac{27j}{j-12^3}}.\label{2}\tag{2}
 \end{equation*}
The relation \eqref{2} is a relation between modular forms, therefore its proof comes for free.
Questions:

Is there an algebraic reason why \eqref{1} should be true?
Is there an algebraic proof of \eqref{1}?
Are there similar formulas for $\Delta^{1/d}$ where $d$ is a divisor of $24$?
Are there similar formulas for $\Delta^{1/d}$ if $d$ is not  a divisor of $24$?
What is the algebraic significance of the number $24$ in this context ($\Delta$ is the $24$-th power of the Dedekind eta function, but what does this mean algebraically)?

REPLY [5 votes]: Regarding Q5, the modular form $\Delta$ has weight 12, so it is a section of the line bundle $\omega^{\otimes 12}$ on the modular curve $Y$. Here $\omega$ is the Hodge bundle, which can be defined as $\omega = \pi_* \Omega^1_{E/Y}$ where $\Omega^1_{E/Y}$ is the sheaf of relative Kähler differentials on the universal elliptic curve $\pi : E \to Y$ (note that here $Y=Y(1)$, so there is no universal elliptic curve, to get around this problem we must treat $Y(1)$ as an algebraic stack).
If $d$ is a divisor of $12$ then $\Delta^{1/d} = \eta^{24/d}$ will be a cusp form of weight $12/d$ (I don't recall the precise level), so everything is algebraic. If you want to get to $\eta$ (in other words $d=24$) then you need half-integral weight modular forms. The interpretation of such modular forms uses the metaplectic cover of $\mathrm{SL}_2(\mathbb{R})$, which is not an algebraic group anymore, only a Lie group (see David Loeffler's answer for more details). The point is that the Hodge bundle $\omega$ is not the square of a line bundle, but it becomes a square once you pull back to the metaplectic group.