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## Is there a nice proof of the fact that there are (p-1)/24 supersingular elliptic curves in characteristic p?

If $k$ is a characteristic $p$ field containing a subfield with $p^2$ elements (e.g., an algebraic closure of $\mathbb{F}_p$), then the number of isomorphism classes of supersingular elliptic curves over $k$ has a formula involving $\lfloor p/12 \rfloor$ and the residue class of $p$ mod 12, described in Chapter V of Silverman's The Arithmetic of Elliptic Curves. If we weight these curves by the reciprocals of the orders of their automorphism groups, we obtain the substantially simpler Eichler-Deuring mass formula: $\frac{p-1}{24}$. For example, when $p=2$, the unique supersingular curve $y^2+y=x^3$ has endomorphisms given by the Hurwitz integers (a maximal order in the quaternions), and its automorphism group is therefore isomorphic to the binary tetrahedral group, which has order 24.

Silverman gives the mass formula as an exercise, and it's pretty easy to derive from the formula in the text. The proof of the complicated formula uses the Legendre form (hence only works away from 2), and the appearance of the $p/12$ boils down to the following two facts:

1. Supersingular values of $\lambda$ are precisely the roots of the Hasse polynomial, which is separable of degree $\frac{p-1}2$.
2. The $\lambda$-line is a 6-fold cover of the $j$-line away from $j=0$ and $j=1728$ (so the roots away from these values give an overcount by a factor of 6).

Question: Is there a proof of the Eichler-Deuring formula in the literature that avoids most of the case analysis, e.g., by using a normal form of representable level?

I suppose any nontrivial level structure will probably require some special treatment for the prime(s) dividing that level. Even so, it would be neat to see any suitably holistic enumeration, in particular, one that doesn't need to single out special $j$-invariants.

(This question has been troubling me for a while, but Greg's question inspired me to actually write it down.)

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In Katz-Mazur there's an elegant uniform proof using the geometry of mod-$p$ modular curves; no mucking around with Legendre model or numerology of 1728, etc. I think it's near the end of some chapter; maybe end of chapter 11 or 12? – BCnrd May 14 2010 at 6:06
This argument goes back to Grothendieck. The only thing that Grothendieck didn't have, which makes a proof of the formula much more natural, is the formula for the degree of a line bundle on a DM-stack in terms of the zero set counted as one does on stacks. – Torsten Ekedahl May 14 2010 at 7:08

One argument (maybe not of the kind you want) is to use the fact that the wt. 2 Eisenstein series on $\Gamma_0(p)$ has constant term (p-1)/24.

More precisely: if $\{E_i\}$ are the s.s. curves, then for each $i,j$, the Hom space $L_{i,j} := Hom(E_i,E_j)$ is a lattice with a quadratic form (the degree of an isogeny), and we can form the corresponding theta series $$\Theta_{i,j} := \sum_{n = 0}^{\infty} r_n(L_{i,j})q^n,$$ where as usual $r_n(L_{i,j})$ denotes the number of elements of degree $n$. These are wt. 2 forms on $\Gamma_0(p)$.

There is a pairing on the $\mathbb Q$-span $X$ of the $E_i$ given by $\langle E_i,E_j\rangle =$ # $Iso(E_i,E_j),$ i.e. $$\langle E_i,E_j\rangle = 0 \text{ if } i \neq j\text{ and equals # }Aut(E_i) \text{ if }i = j,$$ and another formula for $\Theta_{i,j}$ is $$\Theta_{i,j} := 1 + \sum_{n = 1}^{\infty} \langle T_n E_i, E_j\rangle q^n,$$ where $T_n$ is the $n$th Hecke correspondence.

Now write $x := \sum_{j} \frac{1}{\text{#}Aut(E_j)} E_j \in X$. It's easy to see that for any fixed $i$, the value of the pairing $\langle T_n E_i,x\rangle$ is equal to $\sum_{d |n , (p,d) = 1} d$. (This is just the number of $n$-isogenies with source $E_i,$ where the target is counted up to isomorphism.) Now $$\sum_{j} \frac{1}{\text{#}Aut(E_j)} \Theta_{i,j} = (\sum_{j} \frac{1}{\text{#}Aut(E_j)}) + \sum_{n =1}^{\infty} \langle T_n E_i, x\rangle q^n = (\sum_{j}\frac{1}{\text{#}Aut(E_j)}) + \sum_{n = 1}^{\infty} (\sum_{d | n, (p,d) = 1} d)q^n.$$

Now the LHS is modular of wt. 2 on $\Gamma_0(p)$, thus so is the RHS. Since we know all its Fourier coefficients besides the constant term, and they coincide with those of the Eisenstein series, it must be the Eisenstein series. Thus we know its constant term as well, and that gives the mass formula.

(One can replace the geometric aspects of this argument, involving s.s. curves and Hecke correspondences, with pure group theory/automorphic forms: namely the set $\{E_i\}$ is precisely the idele class set of the multiplicative group $D^{\times}$, where $D$ is the quat. alg. over $\mathbb Q$ ramified at $p$ and $\infty$. This formula, writing the Eisenstein series as a sum of theta series, is then a special case of the Seigel--Weil formula, I believe, which in general, when you pass to constant terms, gives mass formulas of the type you asked about.)

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You can also do this "topologically". The idea is to separately calculate the (l-adic etale) Euler characteristics of the stack Y_0(p) in char. 0 and in char. p, then see how they must relate. Here we go:

Over C, and hence over Q_p-bar as well, the Euler characteristic of Y_0(p) is (p+1)*(-1/12), since Y_0(p) is a (p+1)-fold cover of Y=M_{ell}.

On the other hand, over F_p-bar, up to "homeomorphism" Y_0(p) is two copies of Y glued at the supersingular points, so the Euler characteristic is 2*(-1/12) - S (where S is the "number" of supersingular elliptic curves).

However, since Y_0(p) has semi-stable reduction (and is constant at infinity) over Z_p, the special fiber is gotten from the generic fiber by contracting a bunch of "circles" to points, one for each nodal point of the special fiber; thus our second (char p) Euler characteristic is equal to our first (char 0) Euler characteristic plus S. Comparing and solving for S gives the formula pretty quickly.

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 I like this argument. It has a very conversational flavor. – S. Carnahan♦ May 15 2012 at 7:15 Must be all the quotation marks. Anyway, glad you like it. – Dustin Clausen May 15 2012 at 7:17

This will be a summary of the proof referenced by BCnrd in the comments, for the benefit of those that haven't looked through Katz, Mazur, Arithmetic Moduli of Elliptic Curves. A scan is available as an unsearchable djvu near the bottom of Katz's web page. I'd like to point out in particular just how different it is from Emerton's proof, and how the textbook proof I summarized in the statement of the question is essentially the same as this one, but trades some conceptual clarity for simplicity of vocabulary. The relevant corollary (12.4.6) is on page 358, which is page 185 of the scan.

Katz and Mazur start with a moduli problem $P$ (i.e., a functor from $(Ell)$ to Sets) representable by a scheme $M(P)$. We assume that $P$ is defined over an algebraic closure of $\mathbb{F}_p$, and is finite étale over $(Ell/\overline{\mathbb{F}_p})$. This means one has an étale surjection from $M(P)$ to the stack of elliptic curves over $\overline{\mathbb{F}_p}$. There is a distinguished section of the line bundle $\omega_P^{p-1}$, called the Hasse invariant. It is defined as the differential of the Verschiebung, which I will explain a bit later, and it satisfies two key properties:

1. The Hasse invariant vanishes if and only if the curve is supersingular.
2. All zeroes have multiplicity one (Igusa's theorem).

By general line bundle arithmetic, the total number of zeroes of this section in $P$, counting multiplicity, is equal to $p-1$ times the degree of $\omega$, or equivalently, $\frac{p-1}{24}$ times the degree of $P$ over (Ell).

To complete the proof, one shows that the preimage of a point under the composition $M(P) \to \mathcal{M}_{Ell} \to \mathbb{A}^1_j$ has size equal to the degree of $P$ over $(Ell)$ divided by the order of the automorphism group of the underlying elliptic curve, by using representability to deduce the freeness of the group action. This yields $$\deg P \cdot \sum_{\text{supersingular } j} \frac{1}{|\operatorname{Aut} E_j|} = \deg P \cdot \frac{p-1}{24},$$ and we are done.

The Verschiebung and the Hasse invariant deserve some additional explanation. For any $\mathbb{F}_p$-scheme $X$, there is an absolute Frobenius map $X \to X$ which on affines is the functor $\operatorname{Spec}$ applied to the $p$-th power map. For an $\mathbb{F}_p$-scheme $S$ and an $S$-scheme $X$, one obtains an $S$-scheme $X^{(p)}$ as the pullback of $X$ over the absolute Frobenius on $S$. By the universal property of pullbacks, the absolute Frobenius on $X$ factors through an $S$-map $X \to X^{(p)}$, called the relative Frobenius. Over an elliptic curve $E \to S$, this map turns out to be an isogeny of degree $p$. The Verschiebung $V: E^{(p)} \to E$ is defined as the dual to the relative Frobenius isogeny $F: E \to E^{(p)}$, and the multiplication by $p$ map on $E$ factors as $[p] = VF$.

The kernel of Frobenius is always a connected group scheme of length $p$, and the kernel of the Verschiebung over a field is connected if and only if the full $p$-torsion subgroup is connected. The latter case can be taken as a definition of supersingularity over geometric points, and for general $\mathbb{F}_p$-schemes, any family of elliptic curves with at least one supersingular geometric fiber is called supersingular. The Hasse invariant is the induced map on $S$-Lie algebras: $\text{Lie}V: \text{Lie}(E/S)^{(p)} \to \text{Lie}(E/S)$. It is an isomorphism if and only if $E$ is not supersingular (i.e., $E$ is ordinary), as one can calculate by examining formal groups. We can write it as an element of $\text{Hom}(\text{Lie}(E/S)^{(p)}, \text{Lie}(E/S)) = \text{Hom}(\text{Lie}(E/S)^{\otimes p}, \text{Lie}(E/S)) = H^0(S, \omega_{E/S}^{\otimes p-1})$. It is a modular form of weight $p-1$, and its $q$-expansion is identically 1.

The fact that $\omega_P$ has degree $\frac{1}{24}\deg P$ is still a bit of a conceptual mystery to me. It is proved by first calculating it for a full level 3 structure (which has degree 24) and then transferring to all other representable moduli problems.

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One could try to compute the degree of $\omega_P$ in the following way: $\omega_P^{\otimes 12}$ has a section, namely $\Delta$, which has zeroes precisely at the cusps. If we work over $(Ell)$ itself, there is a single cusp where it has a simple zero. But probably one has some phenomena of the following sort: the Tate curve corresponding to the cusp at infinity has automorphisms of order 2, and so the true (i.e. stack theoretic) order of vanishing there is 1/2. Hence deg $\omega^{\otimes 12} = 1/2$, and so deg $\omega = 1/24$. – Emerton Jun 15 2010 at 21:24
Whether or not this is completely correct as written, I think that one should think of the degree of $\omega$ as coming from the existence of the section $\Delta$ which is non-zero at all non-cuspidal points. – Emerton Jun 15 2010 at 21:25
Thanks, that is helpful. I should have thought of $\Delta$. – S. Carnahan Jun 15 2010 at 22:22