66.5Canadian Mathematical Society Conference Proceedings Volume 00, 0000

cmrmn  Which primes are sums of two cubes?  24  FERNANDO RODRIGUEZ VILLEGAS AND DON ZAGIER

cmrmn

Abstract. Let S_{p} be the “unknown” part of the L-series of the elliptic curve x^{3} + y^{3} = p at s = 1, so that conjecturally S_{p} = 0 if p is a sum of two distinct cubes and equals the order of a Tate–Shafarevich group otherwise. The question of the title is then to determine whether S_{p} = 0. For p ≢ 1 (mod 9) the answer depends only on p (mod 9) and is well known. We give three different criteria for the remaining case. Our first formula represents S_{p} as the trace of a certain algebraic number (the value of a specific modular function at a CM point), the second represents S_{p} as the square of the trace of a similar number, and the third shows that S_{p} vanishes if an only if p|f_{2(p−1)/3}(0), where {f_{n}(t)}_{n≥0} is a sequence of polynomials satisfying a simple recursion relation.

1. Introduction and results

A classical problem of Diophantine analysis is to recognize which numbers N are the sum of two rational cubes. For instance, 1 is not so represented (Fermat, Euler), whereas every prime of the form 9k − 1 conjecturally is (Sylvester). If we assume the Birch–Swinnerton-Dyer conjecture, then the question is equivalent to the vanishing at s = 1 of the L-series of the elliptic curve E_{N} : X^{3} + Y^{3} = N.

We consider only the case when N = p is prime. If p ≡ 2, 3 or 5 (mod 9), then L(E_{p}, 1) ≠ 0, so p should not be a sum of two cubes (except for 1^{3} + 1^{3} = 2). This is in fact true and follows either from a 3-descent argument (given already in the 19th century by Sylvester, Lucas and Pepin) or from the Coates-Wiles theorem. If p ≡ 4, 7 or 8 (mod 9) then the functional equation forces L(E_{p}, 1) to vanish, so p should be a sum of two cubes, and for the first two of these three cases a proof of this has been announced by Noam Elkies. From now on we restrict to the remaining case p ≡ 1 (mod 9). Here the L-series may or may not vanish. The question is numerically decidable for any given prime, since

L(E_{p}, 1) = \frac{\sqrt{3} Γ(\frac{1}{3})^{3}}{2π \sqrt[3]{p}} S_{p},

cmrmn2 F. RODRIGUEZ VILLEGAS D. ZAGIER

66.5\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society 0000-0000/00 $1.00 + $.25 per page where S_{p} is known to be an integer (conjecturally equal to 0 if E_{p}(Q) ≠ {0} and to the order of the Tate–Shafarevich group of E_{p} otherwise), but a table of these numbers, such as the one for p < 2000 given at the end of this section, suggests no simple pattern. In this paper we will give three formulas for S_{p} and hence three conjectural answers to the question of the title. Most of the proofs rely on ideas similar to those in [5] and [7] and have been omitted or only sketched, but, to quote from Sylvester’s paper on the same subject [9], “I trust my readers will do me justice to believe that I am in possession of a strict demonstration of all that has been advanced without proof.” We do include a few short proofs which use ideas different from those in the two papers cited.

cmrmn 1991 Mathematics Subject Classification. 11F67, 11D85, 11G40.

First answer: We associate to the prime p = 9k + 1 an algebraic number α_{p} of degree 18k, defined as follows

α_{p} = \frac{\sqrt[3]{p}}{54} \frac{Θ(pδ)}{Θ(δ)} ,

where Θ(z) = \frac{1}{2} ∑_{m,n∈Z} e^{2πi(m^{2}+mn+n^{2})z} and δ = −\frac{1}{2}(1+1/3\sqrt{−3}). (The value of Θ(δ), by the way, is −3Γ(\frac{1}{3})^{3}/(2π)^{2}.) Then

S_{p} = Tr(α_{p}),

where Tr denotes absolute trace. A more detailed statement is given in Theorem 1 below.

Second answer: This has the same form, but gives the square root of S_{p} as a trace (thus proving, in particular, that S_{p} is a square, as expected). Of course S_{p} has two square roots when nonzero. It turns out that they are canonically indexed by the two primes above p in K = Q(\sqrt{−3}). Let P be one of these primes. Choosing P is equivalent, via P = (p, \frac{−r+\sqrt{−3}}{2}), to choosing an integer r (mod 2p) with r^{2} ≡ −3 (mod 4p). Let z_{0} = (r + \sqrt{−3})/2 and set

β_{P} = \begin{matrix}^{6} & \sqrt{p} \\ √ & \overline{±12} \end{matrix} \frac{η(pz_{0})}{η(z_{0}/p)},

where η(z) = e^{πiz/12} ∏_{n=1}^{∞}(1 − e^{2πinz}) is Dedekind’s eta function, the sign is +1 if p ≡ 3 (mod 4) and −1 if p ≡ 1 (mod 4), and the correct choice of the 6^{th} root of p will be explained later. Then β_{P} is algebraic of degree 6k over Q and we have

S_{p} = [Tr(β_{P})]^{2}, and Tr(β\overline{P}) = −Tr(β_{P}).

A more detailed statement is given in Theorem 2 below.

Third answer: Define polynomials f_{n}(t) by f_{0}(t) = 1, f_{1}(t) = t^{2} and

f_{n+1}(t) = (1 − t^{3}) f′_{n}(t) + (2n + 1) t^{2} f_{n}(t) − n^{2}t f_{n−1}(t) (n ≥ 1),

and let A_{k} = f_{3k}(0). (It is trivial that f_{n}(0) = 0 if 3 does not divide n.) Then

S_{p} ≡ (−3)^{\frac{p−10}{3}} (\frac{p − 1}{3})!^{2}A_{2(p−1)/9} (mod p).

This determines S_{p} since |S_{p}| < p/2 as we shall prove in §5. In particular, we have

L(E_{p}, 1) = 0 ⟺ p|A_{2(p−1)/9}.

Third answer (variant): In fact, the A_{k}’s of the third answer (which are normalized central values of certain Hecke L-series) are always squares, and we can get their square roots as follows. Define polynomials g_{n}(t) ∈ Q[t] by g_{0}(t) = 1, g_{1}(t) = \frac{3}{8}t^{2} and

g_{n+1}(t) = (1 − t^{3}) g′_{n}(t) − (2n + \frac{3}{8}) t^{2} g_{n}(t) − n(n − \frac{1}{2}) t g_{n−1}(t) (n ≥ 1),

and let B_{k} = g_{3k}(0). (Just as with f_{n}, g_{n}(0) = 0 if 3 does not divide n.) Then

A_{2k} = B_{k}^{2}, for all k ≥ 0

and in particular

L(E_{p}, 1) = 0 ⟺ p|B_{(p−1)/9},

and

\sqrt{S_{p}} ≡ ±(\sqrt{−3})^{\frac{p−10}{3}} (\frac{p − 1}{3})! B_{(p−1)/9} (mod p).

We conjecture that this formula is always true with the + sign if we interpret \sqrt{S_{p}} and \sqrt{−3} (mod p) as Tr(β_{P}) and r (mod p), respectively (with r, P, β_{P} as in the second answer), and have checked this for p < 2000.

We give a short table of values of the numbers B_{k}.

\begin{matrix}k & B_{k} \\ 0 & 1 \\ 1 & −2 \\ 2 & −152 \\ 3 & −6848 \\ 4 & −8103296 \\ 5 & 22483912960 \\ 6 & −8062284861440 \\ 7 & 196434444070666240 \\ 8 & 532650564250569441280 \\ 9 & 2039228675045199496806400 \\ 10 & −5209573728611533514689740800 \end{matrix}

This also gives the first values of the numbers A_{2k} = B_{k}^{2}. The odd-index values A_{2k+1}, which are not needed for our “third answer,” are also the squares of the constant terms of certain polynomials c_{3k+1}(t) satisfying a recursion (cf. Theorem 3 below, where a short table is given).

The numbers A_{k} and B_{k} have a different description in terms of generating functions:

F (\frac{1}{3}, \frac{1}{3}; \frac{2}{3}; x) = ∑_{k=0}^{∞} \frac{A_{k}}{(3k)!}T^{k}

and

(1 − x)^{\frac{1}{24}} F (\frac{1}{3}, \frac{1}{3}; \frac{2}{3}; x)^{\frac{1}{2}} = ∑_{k=0}^{∞} \frac{B_{k}}{(3k)!} (\frac{−T}{2} )^{k} ,

where F = _{2}F_{1} is Gauss’s hypergeometric function and

T = x \frac{F (\frac{2}{3}, \frac{2}{3}; \frac{4}{3}; x)^{3}}{F (\frac{1}{3}, \frac{1}{3}; \frac{2}{3}; x)^{3}} .

That the coefficients in the first of these hypergeometric expansions are the squares of the coefficients in the second is a surprising and beautiful identity, quite apart from the connection with L-series.

There are similar results for the number S_{p^{2}} corresponding to the elliptic curve E_{p^{2}} : x^{3} + y^{3} = p^{2}, namely

S_{p^{2}} = Tr(γ_{P})^{2}, γ_{P} = \frac{ζ p^{\frac{1}{3}}}{\sqrt{12}} \frac{η(pz_{0})}{η(−\overline{z}_{0}/p)},

for a certain root of unity ζ, and

S_{p^{2}} = 0 ⟺ p|A_{(p−1)/9} ⟺ p|B_{2(p−1)/9}.

However in the rest of this note we will stay with S_{p}.

Using any of the “answers” given in this section, we can easily calculate S_{p} numerically. We give a table for p ≡ 1 (mod 9), p < 2000. In this range, the value of S_{p} is always 0, 1 or 4, as follows:

S_{p} = 0 : 19, 37, 127, 163, 271, 379, 397, 433, 523, 631, 829, 883, 919,
937, 1063, 1171, 1459, 1531, 1567, 1621, 1657, 1801
S_{p} = 1 : 73, 109, 181, 199, 307, 487, 541, 577, 613, 757, 811, 1009,
1117, 1153, 1279, 1297, 1423, 1549, 1693, 1783
S_{p} = 4 : 739, 991, 1747, 1873, 1999

A complete table of S_{N} for N < 1000 is given in [10].

2. The formulas for S_{p}

Let O = Z[ω] ⊂ K = Q(ω) = Q(\sqrt{−3}), where ω^{2} + ω + 1 = 0, \sqrt{−3} = 2ω + 1, and embed K in C via ω ↦ e^{2πi/3}. The elliptic curve E : x^{3}+y^{3} = 1 has complex multiplication by O. Its L-series is L(ψ, s), where ψ is the Hecke character of K satisfying

ψ((α)) = α, for all α ∈ O, α ≡ 1 (mod 3).

Let p ≡ 1 (mod 3) be a prime. We consider the groups

Δ = (O/pO)^{⋆}/(Z/pZ)^{⋆} and Δ_{0} = Δ/O^{⋆},

which are cyclic of orders p − 1 and (p − 1)/3 respectively. We let H_{3p} (H_{p}) be the ring class field modulo 3p (modulo p) of K and identify Δ (Δ_{0}) with Gal(H_{3p}/K) (Gal(H_{p}/K)) via the Artin map.

Let χ : Δ ⟶ ⟨ω⟩ be the cubic character defined by

χ(u) ≡ (\frac{\overline{u}}{u})^{(p−1)/3} (mod p) (u ∈ O, (u, p) = 1).

Then L(ψχ, s) and L(ψχ^{2}, s) are the L-series of the curves E_{p} and E_{p^{2}}, respectively. The sign in their functional equation is +1 if and only if p ≡ 1 (mod 9) or, equivalently, if and only if χ factors through Δ_{0}.

The formulas for S_{p} that we will obtain involve linear combinations of values of certain modular forms on CM points in the upper-half plane corresponding to Δ and Δ_{0}. We need to introduce the following notation in order to do this explicitly.

1) Let δ = (−1−1/3\sqrt{−3})/2 ∈ K∩H, where H denotes the complex upper-half plane. As usual η will denote Dedekind’s eta function. As a set of representatives for Δ we take the numbers 1 and δ − k, with k ∈ Z/pZ such that δ − k is prime to p (hence excluding two values and bringing the total to p − 1).

Let μ_{p} : Δ ⟶ C be given by

μ_{p}(1) = \frac{pΘ(pδ)}{Θ(δ)}, μ_{p}(δ − k) = \frac{Θ(\frac{δ−k}{p})}{Θ(δ)} (k ∈ Z/pZ, (δ − k, p) = 1),

with Θ(z) as in §1. The function μ_{p} is well defined and its values are conjugate algebraic integers in H_{3p}/K.

Note that p^{\frac{1}{3}} ∈ H_{3p} and that p^{\frac{1}{3}} ∈ H_{p} if and only if p ≡ 1 (mod 9). We define κ_{p} = p^{− \frac{2}{3}} μ_{p}(1) = p^{\frac{1}{3}} Θ(pδ)/Θ(δ). It belongs to H_{3p} and its conjugates over K are \{p^{− \frac{2}{3}} \overline{χ}(u)μ_{p}(u) : u ∈ Δ\}.

cmrmn Theorem 1. Let p ≡ 1 (mod 9) be prime. With the above notation we have

Tr_{H_{3p}/K}(κ_{p}) = p^{− \frac{2}{3}} ∑_{u∈Δ} \overline{χ}(u)μ_{p}(u) = 27 S_{p},

where S_{p} ∈ Z is the Birch–Swinnerton-Dyer number defined in the introduction.

2) We choose w a primitive cube of unity modulo p; this corresponds, via P = (w−ω, p), to choosing a prime P of O above p. As a set of representatives for Δ_{0} we take the numbers 1 and ω−k, where k runs over Z/pZ\{w, −1−w, 0, −1})/ ∼, and where ∼, defined by k ∼ −1/(k +1) ∼ −1−1/k, corresponds to orbits under multiplication by ω. (The values of ω − k for k = 0 or −1 represent the same class as 1 in Δ_{0}).

Let λ_{P} : Δ_{0} ⟶ C be given by

λ_{P}(ω − k) = (\frac{ω − k}{P} ) ζ_{24}^{kp} η(\frac{ω − k}{p})/η(ω) (k ∈ Z/pZ, k ≠ w, −1 − w)

λ_{P}(1) = (\frac{2}{p}) ϵ_{p}\sqrt{p} η(pω)/η(ω),

where (\frac{ω − k}{P} ) = (\frac{w − k}{p} ) is the quadratic symbol at P, ζ_{24} = e^{2πi/24}, \sqrt{p} > 0,

and ϵ_{p} = 1, i if p ≡ 1, 3 (mod 4). The function λ_{P} is well defined and its values are conjugate units in an abelian extension of K, which is quadratic over H_{p}.

Let r be a solution of r^{2} ≡ −3 (mod 4p) such that r ≡ 2w + 1 (mod p), and let z_{0} = (r + \sqrt{−3})/2. We define ρ_{P} = (\frac{r}{p}) ζ_{24}^{−p(r+1)/2}η(z_{0}/p)/η(ω). It is not

hard to check that ρ_{P}^{2} ∈ O generates P. Finally, we let ξ_{P} = ζ p^{− \frac{1}{3}} ρ_{P}^{−1}λ_{P}(1), where ζ ∈ O^{⋆} is such that ζ ρ_{P}^{2} ≡ 1 (mod 2); it belongs to H_{p} and its conjugates over K are \{ζp^{− \frac{1}{3}} ρ_{P}^{−1}χ(u)λ_{P}(u) : u ∈ Δ_{0}\}.

cmrmn Theorem 2. Let p ≡ 1 (mod 9) be prime. With the above notation we have

Tr_{H_{p}/K}(ξ_{P}) = p^{− \frac{1}{3}} ρ_{P}^{−1} ∑_{u∈Δ_{0}} χ(u) λ_{P}(u) = \sqrt{−3} R_{P},

with R_{P} ∈ Z. The number R_{P} satisfies

S_{p} = R_{P}^{2} and R\overline{P} = −R_{P},

where S_{p} is the Birch–Swinnerton-Dyer number defined in the introduction.

Remarks. 1. Theorem 2 holds in more generality. For any character ϕ of Δ let

R_{P}(ϕ) = ∑_{u∈Δ} ϕ(u) λ_{P}(u)

(a Lagrange resolvent). Then for ϕ of odd order, R_{P}(ϕ)^{2} is essentially the algebraic part of L(ψϕ^{−2}, 1).

2. It is possible to define, in a similar way, integers R_{A} associated to any cubefree ideal A of O such that R_{A}^{2} = S_{N} is the Birch–Swinnerton-Dyer number of the elliptic curve x^{3} + y^{3} = N, where N = N(A). One might hope that these numbers are the Fourier coefficients of a modular form of some sort.

Theorem 1 is proved by writing the special value L(ψχ, 1) as a linear combination of values of an Eisenstein series as in [4] and using the Shimura reciprocity law. One then deduces Theorem 2 from a variant of the factorization formula of [7] and a careful chasing of 24^{th} roots of unity. Theorems 1 and 2 are easily seen to be equivalent to the analogous statements given in the introduction.

3. Congruences

Our third answer to the question when S_{p} vanishes was based on a congruence between S_{p} (which is, up to a factor, the value of a certain L-function at s = 1) and another number A_{k} which, as we will discuss in a moment, is (again up to a factor) a special value of an L-function independent of p at some other value of s. Congruences of this sort go back to Cauchy, Kummer, and Hurwitz. For example [1], the class number h(−p) of the quadratic field Q(\sqrt{−p}) for a prime p > 3, p ≡ 3 (mod 4) satisfies h(−p) ≡ −2B_{(p+1)/2} (mod p), where (here only!) B_{n} denotes the n^{th} Bernoulli number. One way to interpret this is to say that the two Dirichlet series

∑_{n≥1} (\frac{n}{p}) n^{−s}  and  ∑_{n≥1} n^{\frac{p−1}{2}} n^{−s} ,

which are congruent term by term modulo p, also have congruent values at s = 0. Turning this fact into a heuristic argument, we would expect that (suitable algebraic versions of) the values at s = 0 of the Dirichlet series

∑_{α∈O\\{0\}} χ(α) \frac{1}{\overline{ψ}(α)N(α)^{s}} and ∑_{α∈O\\{0\}} ( \frac{α}{\overline{α}} )^{\frac{2(p−1)}{3}} \frac{1}{\overline{ψ}(α)N(α)^{s}} ,

which are easily seen to equal L(ψχ, 1) and L(ψ^{2k−1}, k) respectively, with k−1 = 2(p − 1)/3, should also be congruent modulo p (see §2 for notations). This is indeed the case, at least if p ≡ 1 (mod 3), where it follows from the existence of a p-adic L-function interpolating special values of Hecke L-series due to Manin– Vishik and Katz. We now make this precise.

For k ∈ N define the algebraic part of L(ψ^{2k−1}, k) to be

L_{k} = 3ν (\frac{2π}{3\sqrt{3} Ω^{2}} )^{k−1} \frac{(k − 1)!}{Ω} L(ψ^{2k−1}, k),

where ν = 2 if k ≡ 2 (mod 6) and ν = 1 otherwise, and Ω = Γ(1/3)^{3}/(2π\sqrt{3}) = 1.766638 ⋯ is the fundamental real period of the elliptic curve x^{3} + y^{3} = 1. Then using the formulas of [2] we find that

S_{p} ≡ (−3)^{\frac{p−10}{3}} (\frac{p − 1}{3})!^{2}L_{(2p+1)/3} (mod p).

This corresponds to the “third answer” of the introduction because L_{3n+1} = A_{n} for all n, as we will now discuss.

4. Recursions

By the methods of [7] one can obtain formulas for L(ψ^{2k−1}, k), k ∈ N, in terms of derivatives of modular forms and then deduce recursive formulas giving the algebraic parts L_{k}, and similarly for their square roots. A typical formula for the values is the identity L_{3n+1} = A_{n} just mentioned, but since the results for the square roots are more interesting and give more precise information, we shall state the full results only there (but see Remark 2 after Theorem 4 below).

The formulas for the square roots of L_{k} can be divided naturally into three branches: (a) k ≡ 1 (mod 6), (b) k ≡ 2 (mod 6), and (c) k ≡ 4 (mod 6). (For other values of k the functional equation forces L(ψ^{2k−1}, k) to be zero.) For each branch there is a formula for \sqrt{L_{k}} in terms of a higher non-holomorphic derivative of a fixed half-integral weight modular form at a fixed CM point (e.g. for \sqrt{L_{6n+1}} it is the n^{th} non-holomorphic derivative of η(z) at the point z = ω), and this in turn leads to the following description of the square roots as the constant terms of a recursively defined sequence of polynomials.

cmrmn Theorem 3. Let a_{n}(t), b_{n}(t), c_{n}(t) be the polynomials defined by the recursions

\begin{aligned}a_{n+1}(t) & = −(1 − 8t^{3}) a′_{n}(t) − (16n + 3) t^{2} a_{n}(t) − 4n(2n − 1) t a_{n−1}(t), \\ b_{n+1}(t) & = (1 − 8t^{3}) b′_{n}(t) + (16n + 9) t^{2} b_{n}(t) − 4n(2n + 1) t b_{n−1}(t), \\ c_{n+1}(t) & = (1 − 8t^{3}) c′_{n}(t) + (16n + 9) t^{2} c_{n}(t) − 4n(2n + 1) t c_{n−1}(t) \end{aligned}

for n ≥ 1, with initial conditions

a_{0}(t) = 1, a_{1}(t) = −3t^{2}, b_{0}(t) = 1, b_{1}(t) = 9t^{2}, c_{0}(t) = t, c_{1}(t) = 1 + t^{3}.

Then for all n ∈ Z_{≥0}

L_{6n+1} = a_{3n}(0)^{2}, L_{6n+2} = b_{3n}(0)^{2}, L_{6n+4} = c_{3n+1}(0)^{2},

while a_{m}(0) = b_{m}(0) = c_{m+1}(0) = 0 for m ≢ 0 (mod 3).

We give a short table. Note that a_{3n}(0) is the B_{n} of the introduction.

\begin{matrix}n & a_{3n}(0) & b_{3n}(0) & c_{3n+1}(0) \\ 0 & 1 & 1 & 1 \\ 1 & −2 & 6 & −8 \\ 2 & −152 & −216 & 1240 \\ 3 & −6848 & −119232 & −621440 \\ 4 & −8103296 & 24105600 & −51596800 \end{matrix}

Remark. The constant terms of the polynomials ∗_{n}(t) (∗ = a, b or c) satisfy the congruences ∗_{n+(p−1)/2}(0) ≡ (π + \overline{π}) ∗_{n} (0) (mod p) for all n > 1 and all p ≡ 1 (mod 3) prime, where π denotes a generator of a prime in K above p with π ≡ 1 (mod 3). The corresponding congruence for the squares of the ∗_{n}(0) (i.e., for the numbers L_{k}) was known, and the possibility of choosing the signs in such a way that this congruence descends to the square roots for all p simultaneously had been conjectured by Koblitz [3]. In fact, Koblitz conjectured the existence of a p-adic L-function interpolating suitable modifications of the presumed square roots. This has been proved by Sofer [8] in other similar cases (see also [6]).

There is another way of obtaining the numbers a_{n}(0), b_{n}(0) and c_{n}(0) directly in terms of generating series.

cmrmn Theorem 4. (1) Let u(τ) = F (\frac{1}{3}, \frac{1}{3}; \frac{2}{3}; τ^{3}) and v(τ) = τ F (\frac{2}{3}, \frac{2}{3}; \frac{4}{3}; τ^{3}), where F = _{2}F_{1} is Gauss’s hypergeometric function. Let

h_{a}(τ) = u(τ)^{1/2}(1 − τ^{3})^{1/24}, h_{b}(τ) = h_{a}(τ)^{3}, h_{c}(τ) = \frac{1}{2}τ h_{b}(τ)

and define

H_{a}(x) = h_{a}(τ), H_{b}(x) = h_{b}(τ), H_{c}(x) = h_{c}(τ),

where x = v/2u = \frac{1}{2}(τ + \frac{1}{6}τ^{4} + \frac{103}{1260}τ^{7} + \frac{169}{3240}τ^{10} + ⋯). Then

H_{a}(x) = ∑_{n≥0}(−1)^{n}a_{n}(0)\frac{x^{n}}{n!} , H_{b}(x) = ∑_{n≥0} b_{n}(0)\frac{x^{n}}{n!} , H_{c}(x) = ∑_{n≥0} c_{n}(0)\frac{x^{n}}{n!} .

(2) The series H_{a}(x) is the expansion of η(z) about ω = (−1 + \sqrt{−3})/2 in the following sense:

(1 − x)^{− \frac{1}{2}} η (\frac{ω − \overline{ω}x}{1 − x}) = c_{1}H_{a}(c_{2}x), (|x| < 1),

where c_{1} = η(ω) = e^{\frac{2πi}{48}} (3^{\frac{1}{4}} Ω/2π)^{\frac{1}{2}} and c_{2} = −3\sqrt{3} Ω^{2}/4π with Ω as in §2.

Remarks. 1. As a corollary of the identity H_{b}(x) = H_{a}(x)^{3} we obtain somewhat surprising polynomial relations between the square roots of the L-values {L_{6n+1}} and {L_{6n+2}}. Analogous identities also hold for other CM curves, linking L-values of one curve to L-values of its twist by Q(\sqrt{−3})/Q; ultimately they boil down to classical Jacobi identity θ′(0) = 2πη^{3}.

2. We briefly state here the power series expansions involving the L_{k} themselves. With the notation of Theorem 4 let G_{0}(x) = u(τ) and G_{1}(x) = τ u(τ). Then 2^{−n}G_{0}^{(n)}(0) equals a_{n/2}(0)^{2} if n ≡ 0 (mod 6), c_{(n−1)/2}(0)^{2} if n ≡ 3 (mod 6), and 0 otherwise; 2^{−n}G_{1}^{(n)}(0) equals b_{(n−1)/2}(0)^{2} if n ≡ 1 (mod 6) and zero otherwise. We may even separate the two branches (a) and (b) in G_{0} by considering the series u(τ)(1±(1−τ^{3})^{\frac{1}{3}} ). It is presumably possible to prove directly that the series H_{a}, H_{b}, H_{c}, G_{0} and G_{1} have their Taylor coefficients related as indicated, but we have not done so.

Proof (sketch). Part 1) of the theorem follows from part 2) and the interpretation of the constant terms a_{n}(0), b_{n}(0), c_{n}(0) as non-holomorphic derivatives of holomorphic modular forms, together with the general fact that the expansion of any modular form as a power series in a modular function satisfies a linear differential equation. (Classical examples of this latter assertion are the expansion of η^{2} or \sqrt[4]{E_{4}} as a power series in 1/j or of θ^{2} as a power series in λ, all of which involve hypergeometric functions.) It can also be proved directly from the recursive definitions of the polynomials a_{n}, b_{n} and c_{n} without any a priori knowledge that modular forms are involved.

Part 2) is a consequence of the following simple result about non-holomorphic derivatives. We recall their definition. For any k ∈ R we let ϑ_{k} be the differential operator \frac{∂}{∂z} + \frac{k}{2iy} acting on functions f of z = x + iy ∈ H. It has the property

(ϑ|_{k+2})γ = ϑ(f|_{k}γ), (γ ∈ Sl_{2}(R)),

where |_{k} has the usual meaning. In particular, if f is a modular form of weight k on some group Γ ⊂ Sl_{2}(R), then ϑ_{k}^{n}f, where ϑ_{k}^{n} = ϑ_{k+2n} ∘ ⋯ ∘ ϑ_{k+2} ∘ ϑ_{k}, is a (non-holomorphic) modular form of weight k + 2n on the same group.

cmrmn Proposition 1. Let f : H ⟶ C be an analytic funtion and z_{0} = x_{0} + iy_{0} a point in H. Then the following expansion holds

∑_{n≥0} ϑ_{k}^{n}f(z_{0})\frac{(2iy_{0}w)^{n}}{n!} = (1 − w)^{−k}f (\frac{z_{0} − \overline{z}_{0}w}{1 − w} ) (|w| < 1).

Proof. It is easy to check by induction that

\frac{1}{n!}ϑ_{k}^{n}f(z_{0}) = ∑_{j+l=n} \binom{j + l + k − 1}{l} (\frac{1}{2iy_{0}} )^{l} \frac{f^{(j)}(z_{0})}{j!}.

Hence,

∑_{n≥0} ϑ_{k}^{n}f(z_{0})\frac{(2iy_{0}w)^{n}}{n!} = ∑_{j≥0} \frac{f^{(j)}(z_{0})}{j!}(2iy_{0}w)^{j} ∑_{l≥0} \binom{j + l + k − 1}{l} w^{l},

and our claim follows from Taylor’s and the binomial theorems.

Remark. Notice that the substitution ϕ(t) = (z_{0}−\overline{z}_{0}t)/(1−t) is an isomorphism from H to the unit disk sending z_{0} to 0, with inverse ϕ^{−1}(z) = (z − z_{0})/(z − \overline{z}_{0}). The proposition then says that the non-holomorphic derivatives ϑ_{k}^{n}f(z_{0}) are essentially the Taylor coefficients of f|_{k}ϕ at t = 0.

5. Estimates of S_{p}

The last two criteria for the vanishing of S_{p} described in the introduction are given in terms of the vanishing of S_{p} (mod p). That these two statements are in fact equivalent is a consequence of the following general estimate.

cmrmn Proposition 2. Let E/Q be a modular elliptic curve of conductor N. Then

|L(E, 1)| < (4N)^{1/4} (log \frac{\sqrt{N}}{8π} + γ) + c_{0} ,

where γ = 0.577 … is Euler’s constant and c_{0} = ζ(\frac{1}{2})^{2} = 2.13263 . . ..

Proof. Because of the universal estimate |a_{n}| ≤ \sqrt{n} σ_{0}(n) for the coefficients of L(E, s) we have

|L(E, 1)| = |(1 + w) ∑_{n=1}^{∞} \frac{a_{n}}{n} e^{−2πn/\sqrt{N}} | ≤ 2 F (\frac{2π}{\sqrt{N}} ) ,

where w = ±1 is the sign in the functional equation and F (x) = ∑_{n=1}^{∞} \frac{σ_{0}(n)}{\sqrt{n}} e^{−nx}.

Using the fact that the Mellin transform of F (x) is Γ(s)ζ(s + \frac{1}{2})^{2}, we find the asymptotic expansion

F (x) = \sqrt{\frac{π}{x}} (log \frac{1}{4x} + γ) + ∑_{n=0}^{∞} c_{n}x^{n} (x ↘ 0)

with c_{0} = ζ(\frac{1}{2})^{2} and c_{1} = −ζ(−\frac{1}{2}) < 0. Some numerical work shows that ∑_{n≥1} c_{n}x^{n} < 0 for all x > 0. (For the proposition, we need this only for x ≤ 2π/\sqrt{11}.)

Applying this to the curve E_{p}, whose conductor is 27p^{2}, we find after a simple calculation the following estimate.

cmrmn Corollary. For p ≡ 1 (mod 9) we have

|S_{p}| < 0.61 p^{5/6} log p .

In particular, S_{p} is determined by its value modulo p and

S_{p} ≡ 0 (mod p) ⟺ S_{p} = 0.

Remark. We can also estimate S_{p} using Theorems 1 or 2. For instance, from Theorem 2 and the estimate η(\frac{ω−k}{p} ) = O(p^{1/4}), we obtain R_{P} = O(p^{2/3}), so that R_{P}, and hence S_{p}, is determined by its value modulo p. The corresponding estimate using Theorem 1 is more difficult, because Θ is not a cusp form, but seems to lead to the estimate S_{p} = O(p^{5/6+ε}), essentially the same as in the Corollary above. Note that to determine S_{p} from its value modulo p we need only the weaker estimate S_{p} < p^{2}/4, since we know that S_{p} is a square.

mReferences

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Department of Mathematics, Princeton University, Princeton, NJ 08544, USA E-mail address: villegas@math.princeton.edu

Max-Planck-Institut für Mathematik, G.-Claren-Str. 26, 53225 Bonn, Germany E-mail address: zagier@mpim-bonn.mpg.de