In the CM case one knows by Deuring [Deu] that the density of supersingular primes is 1/2. Elkies in 1987 [Elk1], when he demonstrated that there are an infinite number of supersingular primes for any elliptic curve E/Q.
A refinement of strong multiplicity one
In section 2.1 we provide some background information on the Sato-Tate conjecture, which is a crucial part of our heuristics. In section 2.4 we discuss the results when applied to certain examples of number fields, and finally in section 2.5 we explain how we can refine the averaging, using a Fouvry-Murty idea.
The heuristic
- Setup
- Approximation model
- Asymptotic Behaviour
- Investigating F (t 0 )
Let k be the order of the torsion in the Mordell-Weil group of the elliptic curve. If we have incompatibility between the congruence arising from torque injection and the congruence satisfied by the norms of the elements of R, then the set Q = {p ∈ R | Np+ 1 ≡ 0 (modk)} can only have a finite number of elements (as in the example mentioned in the introduction).
Proof of the conjecture on average
Setup
However, later in this chapter we shall see that, on average, there is actually a bias in the proportion of supersingular primes, reflected in the value of the constant, which is contrary to the expectations of the heuristic. We will later choose some examples of abelian expansions and use congruence conditions to determine the set of rational primes below the degree one prime of the expansion.
Lifting supersingular elliptic curves
- The case f = 1
- The case f = 2
Since every Fp has a nontrivial square root of unity, the affine equations are exactly the same as t = ±t0, so the magnitude of the isomorphism class is (p−1)/2. In the case of p6≡1 (mod 4), we note that if there were (correct) fourth roots of unity, they would generate an order 4 subgroup of F×p. So there are no fourth roots of unity and so t4=t04 ifft=±t0, so again we must have (p−1)/2 the size of our isomorphism class.
Note that there are no common factors for n0 and m0, otherwise they would also divide k, which would contradict the prime decomposition of k. To interpret this, we consider the possible values of the divisors lk and gls of l (remember that we have l = lkls). In the first case, we note that prime numbers p, such that p|k and p-m/k are exactly the prime numbers that can divide
Note that we must use −b ≡c(k) in the second sum so that the two initial congruence conditions are compatible. We consider the second term on the right-hand side of (2.20) and using Cauchy-Schwarz bound it with We put all of the above together to get that the sum in the statement of the lemma is equal to ϕ(4n)/ϕ(4k) if and only if ls is a square.
If it is divisible by 2 only once, i.e. if we have a given congruence condition of the form p≡c(m), where m = 2m0 form0 is odd, then this can be decomposed into the conditions p≡c(m0) and p≡1(2 ) (since (c, m) = 1 by assumption) and so the second condition can be essentially ignored - all it does is exclude the prime number 2 and thus has no effect on the asymptotic expression. The second expression above can be constrained using Cauchy-Schwarz and Gallagher's theorem as before to give this.
The value of K P
Applications
Imaginary quadratic fields
Real quadratic fields
Cyclotomic fields
The sum of the CP coefficients modulo 15 of the different congruence relations is π/3, and thus the average value is π/24, which is larger than the coefficient in the above asymptotics. So again, we seem to have a smaller than average occurrence of supersingular primes that are completely divisible in the number field. The bias in the distribution of supersingular primes, as seen in these examples, can be traced to the L functions of Section 2.3.
Consider the inner sum, for example for f = 1, of the right-hand side of equation (2.4) and express it using Euler products. This suggests that choosing a sequence of prime numbers such as Q leads to a greater constant in the averaging expression. We also note that the ratio between the two factors is larger (and thus the bias is larger) when q is a smaller prime number.
A refined averaging
We plan to investigate elsewhere the phenomenon of conformal class bias for individual elliptic curves, rather than averaged over a family of elliptic curves, as we have done here, in part by scrutinizing modp Galois representations. In this chapter, we construct (arbitrarily) thin families of elliptic curves for which, on average, the Lang-Trotter conjecture for averaging the Weierstrass curve equations holds.
Weierstrass equations
Note that given such a sequence {an}, we have sequences {an+c} for a fixed integer c that also satisfy the above conditions. Moreover, the sequences {kan+c} for invariant integers inc satisfy the above conditions for all but finitely many primes (ie, those that are sopranomeric with k). So let S={Ei,j :y2=x3+aix2+aj} be our family of elliptic curves and note that if (n) is a polynomial of the second degree or an exponential function, then S is a thinner family than any of those mentioned in the first paragraph of this section.
We will determine that this family meets the Lang-Trotter conjecture on average, using Fouvry-Murty [FM] techniques.
Proof
First sum
We now determine that the same estimates hold if we restrict the inner sum to prime values of n (instead of powers of primes). This is dominated by the xV /(logx)c term and therefore does not affect the asymptotic.
Second sum
The asymptotic
For a given number field F, let A0(GLn(AF)) be the set of cupidal automorphic representations π = ⊗0vπv of GLn(AF). For π ∈ A0(GLn(AF)) for any place v in F, where π is unbranched, we denote the Langlands conjugacy class by A(πv) ⊂GLn(C), which will be represented by the diagonal matrix diag{α1, v, α2, v,. We believe that both Theorems 4.1 and 4.2 could be extended with some caution to the case of cupidal automorphic representations for GL(2) with nontrivial unitary centrality.
One also knows that Sym2π is cuspidal if π is non-dihedral [GJ] and that Sym3π is cuspidal if π is neither dihedral nor tetrahedral [KS1]. One also knows that L(s, π0) does not vanish on Re(s) = 1 for any cuspid automorphic representationπ0 forGLn(AF) [JS2]. Specifically, one can consider LT(s,Sym3π×π) (where T is defined as above) and note that since Sym3π is cuspid, this incomplete Rankin-Selberg L-function does not have a pole ats= 1.
Since π has trivial character, we have thatα1α2 = 1 and therefore we will use the notation α, α−1 for α1, α2, respectively. Since (as mentioned) the incomplete L-function on the left has no pole at s= 1 and the incomplete L-function associated with the symmetric square is non-zero, we have that LT(s,Sym4π) has no pole does not have at s= 1. Further note that since the symmetric third and fourth powers of π are known as isobaric sums of unit cuspid automorphic representations, by Jacquet-Shalika [JS1].
Proof of theorem 4.1 in the non-tetrahedral case
The purpose of the above lemma is to be able to determine the asymptotic behavior of functions L terminated as s → 1+. Note that it suffices to compare incomplete functions L since the number of branching and infinite places is finite, and so the incomplete function L associated with finite unbranched places has the same type of pole at s= 1 as the function Full L. . Considering the behavior of incomplete functions L, for π, π0 that are not dihedral or tetrahedral, as real s→1+, we get
Proof of theorem 4.2 in the non-tetrahedral case
Since π is a cuspidal automorphic representation that is neither dihedral nor tetrahedral, we know that the first L-function on the right has no pole, and the cuspidality of π tells us that the second L-function on the right also has no pole.
An example
In the context of Theorem 4.1, we note that since the upper automorphic representations ρ, ρ0 are locally isomorphic to a set of density sites and thus of lower density greater than 3/5, the representations must have symmetric squares that are twist-equivalent.
The tetrahedral case
Given that Sym2 is a GL(3) cusp shape with a trivial character, the corresponding L function has no pole ats= 1 and is also non-zero at that point. We also know that the Rankin-SelbergL function on the left side of the L-function identity from the lemma has a simple pole at s= 1. Examining the identity from the lemma, we obtain that the symmetric fourth power L - function has no pool ats= 1 and is non-zero at that point.
We also note, by Jacquet-Shalika [JS1] and Shahidi [Sha], that the L-function associated with the symmetric fourth power is non-vanishing ats = 1, since the automorphic representation of the symmetric fourth power is an isobaric sum of unitary cuspidal automorphic representations. Let π, π0∈ A0(GL2(AF)) be tetrahedral representations with a trivial central character and symmetric squares that are not twist equivalent.
The tetrahedral case: an example
- The binary tetrahedral group
- Irreducible representations
- Galois structure
- Combined Galois structures for two different tetrahedral Artin
- The compositum
- Counting
- Symmetric squares
We have three one-dimensional irreducible representations that arise from the irreducible representations on C, leaving us with four irreducible representations whose degrees are nigiveP. Starting with the one-dimensional representations, we note that it is based on C, where we assign the generator of this subgroup to 1,ζ or ζ2, where ζ is the primitive root of unitye2πi/3. Note that F/Q is Galois since {±1}CAf4 and moreover the Galois group is isomorphic to A4.
Note that we could also require F1 = F2, which gives us a representation with 5/8 of the common Frobenius traces (assuming that K1 6=K2), but they will have symmetric turn-equivalent squares, which which is not what we want. Thus, not only do we need to determine the degree of the composite extension, but also its Galois group. Given a conjugacy class of this Galois group, if we take the trace of the individual components and observe them to be equal, then this will correspond to a set of rational primes P of given positive density, where trρ1(Frobp) = trρ2(Frobp).
Since we can cancel such representations to Gal(Q/Q), we obtain with Arthur-Clozel [AC] the existence of tetrahedral cuspidal automorphic representationsπ, π0 such that the set {v|a(πv) =a(π0v)} a density of 17/32. We must now establish that the two representations do not have twist-equivalent symmetric squares. Simple Algebras, Base Change, and the Advanced Theory of the Trace Formula, Volume 120 of Annals of Mathematics Studies.
