Why is the supersingular locus the zero locus of a modular form? This question is related to my other question here: Examples of subspaces singled out by modular forms.
Here I am wondering if there is a philosophical explanation about why the supersingular locus on the modular curve mod p is the zero locus of a modular form.
Even better, are there sufficient conditions ensuring that a subset of the modular curve is the zero locus of a modular form?
Thank you for your help.
 A: On one hand, supersingularity is a "modular" property, i.e. a property of the elliptic curve represented by the point on the modular curve, so it should have a modular description. On the other hand, any finite set of $j$-values $j_1,\ldots,j_n$ is the set of zeros of a modular form. For example: $\Delta^n\prod(j-j_i)$.
A: I'm not sure I buy Felipe Voloch's answer, since there are many choices for his modular form; however, there is a natural modular form cutting out the supersingular locus.  This is something special about elliptic curves; in general (e.g. for surfaces) there is not a known scheme structure on the supersingular locus, at least to my knowledge.
Let $X=X(N)$ be a modular curve with $N$ prime to the characteristic $p$, and let $\pi:  E\to X$ be the universal curve.  Then there is a natural map of line bundles $$f: R^1\pi_*F^*\mathcal{O}_E\to R^1\pi_*\mathcal{O}_E$$ where $F$ is Frobenius.  By cohomology and base change, the supersingular locus is exactly where this map vanishes (that is, an elliptic curve is supersingular if Frobenius acting on $H^1(\mathcal{O})$ is $0$).  Now $\mathcal{O}_E\simeq T_{E/X}$ (slightly non-canonically) using the group structure on $E$, so $R^1\pi_*\mathcal{O}_E$ is $T_X$ (using the moduli description of $X$; $H^1(T_E)$ classifies deformations of $E$).  Similarly $$R^1\pi_*F^*\mathcal{O}_E=F^*R^1\pi_*\mathcal{O}_E=T_X^{\otimes p}.$$  So we may view $f$ as a map $$f: T^{\otimes p}_X\to T_X,$$ or in other words a section to $\omega_X^{p-1}$, also known as a modular form of weight $2p-2$.
Hopefully I got that all right; if not, perhaps an expert can correct me.
