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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.

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2 Answers 2

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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: $\Delta^n\prod(j-j_i)$.

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  • $\begingroup$ Thank you for your answer, but I had a more general situation in mind. Let me try to explain myself better. Consider a property of elliptic curves P, so "modular" as you say, then is there a way to know if the points of the modular curve that satisfy P is the zero locus of a modular form? $\endgroup$
    – Bear
    Commented Sep 6, 2014 at 19:28
  • $\begingroup$ And relating this to the supersingular locus. If you didn't know that there are only finitely many isomorphism classes of supersingular elliptic curves, could you explain why there exists the Hasse invariant that cuts out the supersingular locus? $\endgroup$
    – Bear
    Commented Sep 6, 2014 at 19:33
  • $\begingroup$ I just proved that every finite set of points is the zero locus of a modular form. The supersingularity condition is clearly closed, so either finite or everything. $\endgroup$
    – Felipe Voloch
    Commented Sep 6, 2014 at 19:37
  • $\begingroup$ Ok, you know that the supersingular condition is closed because you studied that there exists the Hasse invariant. However, how could you have guessed that the Hasse invariant existed without knowing it? $\endgroup$
    – Bear
    Commented Sep 6, 2014 at 19:53
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    $\begingroup$ Another way. If an elliptic curve over a base has a non-zero point of order $p$, then on an open set of the base, this point specializes to a non-zero point, still of order $p$, so ordinarity is open. $\endgroup$
    – Felipe Voloch
    Commented Sep 6, 2014 at 19:59
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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.

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  • $\begingroup$ (I've ignored all twists by cusps here, which are necessary to make this correct, I think.) $\endgroup$
    – Daniel Litt
    Commented Sep 6, 2014 at 18:15
  • $\begingroup$ The Hasse invariant is a modular form of weight $p-1$ vanishing on the supersingular locus. This is basically what you are doing. The issue is the formulation of the question, really. $\endgroup$
    – Felipe Voloch
    Commented Sep 6, 2014 at 18:23
  • $\begingroup$ @FelipeVoloch: Yes, I'm making the Hasse invariant essentially; my only complaint is that your construction would for example let one single out the curve of $j$-invariant $5$, whereas the Hasse invariant is somewhat more canonical. Your answer is of course perfect for the question as stated. $\endgroup$
    – Daniel Litt
    Commented Sep 6, 2014 at 18:36

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