Diagonal hypersurfaces and the Bloch-Kato conjecture Sir Peter Swinnerton-Dyer

1.Introduction The research reported in this paper reflects a desire to study the L-function associated with the nonsingular hypersurface W = W_{m,n} in P^{m} defined over Q which is given by the equation

a_{0}X_{0}^{n} + … + a_{m}X_{m}^{n} = 0  (1)

where m > 1 and n > 2. Here we can take the a_{i} to be nonzero integers. The only such L-functions which are interesting are those in middle dimension. The behaviour of these at integral values of s is the subject of the Tate conjecture and the Bloch-Kato conjecture. Though both of these were originally stated about varieties, it appears that they are both really about motives.

Let V be a complete nonsingular variety defined over Q. The outward and visible sign of a motive associated with V is a subspace of the cohomology space which is mapped into itself by every Frobenius map. Hence there is an L-series attached to any motive, and any L-series belonging to V is the product of the L-series belonging to the irreducible motives associated with V in the appropriate codimension. We call a motive geometric if the corresponding cohomology space H_{geom} is spanned by the classes of subvarieties of V defined over Q̄, and transcendental if the corresponding cohomology subspace has trivial intersection with H_{geom}. For any such V, the Tate conjecture primarily concerns the geometric motives; their L-series are believed to be multiplicative combinations of the zeta-functions of algebraic number fields. The Bloch-Kato conjecture describes both sorts of motive, but should probably be split up according as the motive is geometric or transcendental. The transcendental part is the more novel.

In this paragraph, let L(s) be any motivic L-function and d the associated codimension. It is known that the series for L(s) is absolutely convergent in ℜs > \frac{1}{2}d + 1; and it is conjectured that L(s) can be meromorphically continued to the whole s-plane and that there is a functional equation connecting L(s) and L(d+1−s). According to the Bloch-Kato conjecture, if s > \frac{1}{2}(d+1) is in Z then there is a fairly well defined transcendental T such that L(s)/T is in Q; moreover

L(s) = T F G

where G is the order of a certain ètale cohomology group and F (the fudge factor) is a product of Tamagawa numbers. There is a finite set of integer values of s, called strategic values, for which the conjectured formula for T is less complicated than it is in general. (For a geometric motive d has to be even, and the only strategic value is s = \frac{1}{2}d + 1.). A modified version of the Bloch-Kato conjecture, for when d is odd and s = \frac{1}{2}(d + 1), takes some account of the fact that L(s) may vanish there; this appears to be more or less compatible with the Birch/Swinnerton-Dyer conjecture. But for the latter conjecture, when L(\frac{1}{2}(d+1)) is non-zero F is the product of Tamagawa numbers with a certain non-local factor. It is in general difficult to obtain any arithmetic information about the order of a cohomology group. The Tate-Safarevic group is an exception, but this is because we also have a low-brow description of it. It would be good to have similar low-brow descriptions of the groups which occur in the Bloch-Kato conjecture.

The motivic L-functions associated with hypersurfaces of the form (1) are Hecke L-functions. The object of this paper is to compute and interpret L(s) at strategic values of s for transcendental motives of such hypersurfaces; the same methods can be used to compute L(s) at non-strategic values of s, but I am not able to interpret these. I have computed L(s) only for the values n = 4, 5 and 10, but I have no reason to doubt that these are typical. The numerical evidence implies that L(s) = T R where T is a transcendental which can be exactly specified and R is a rational number whose denominator is a power of 2; when n = 4 this can probably be proved by arguments like those in [2] but considerably more complicated. When n = 5 or 10, the rationality of R can presumably again be proved by means of the theory of complex multiplication as developed by Shimura and his pupils; but any statement about the denominator of R would depend on considerable advances in that theory.

The case when d is odd and s = \frac{1}{2}(d+1) is of particular interest, because it is then plausible to hope that the G in the Bloch-Kato conjecture should be a square. Indeed, in all the examples which I have computed R is the product of a square and a power of 2. (A similar phenomenon, though associated with a quite different variety, was already reported in [4].) In the main table contained in this paper, there are two examples where L(s) has a zero of positive even order. The whole behaviour is so similar to what happens in the particular case described by the Birch/Swinnerton-Dyer conjecture that one is bound to wonder whether in this more general situation there is an Abelian variety with a role to play. Moreover there is an obvious Abelian variety in the wings.

Question 1 Does the intermediate Jacobian play a role in the theory?

Again, there is one obvious reason why the order of a finite abelian group might be compelled to be a square: this is if it admits a non-singular skewsymmetric bilinear form. It is well-known that this happens for the Tate-Safarevic group, the bilinear form being the Cassels form.

Question 2 When s = \frac{1}{2}(d + 1), is there a natural skew-symmetric bilinear form on the cohomology group appearing in the Bloch-Kato conjecture?

I am deeply indebted to John Tate for a great deal of helpful advice, without which this paper could not have been written.

2. Preliminaries For any n > 2, let ζ_{n} = e^{2πi/n} and write k = Q(ζ_{n}) with ring of integers 𝔬. Let ρ = ζ^{2} if 2||n, and ρ = ζ otherwise. Let p be a prime of Z which does not divide n or any of the a_{i}, so that W has good reduction W̃ at p. Let 𝔭 be a prime factor of p in 𝔬. Write q = p^{ν} = Norm(𝔭) and fix an isomorphism 𝔬/𝔭 ≈ F_{q}. Thus q is the least positive power of p which is congruent to 1 mod n.

To any nonzero element x in F_{q} there corresponds a unique n^{th} root of unity ϵ whose reduction mod 𝔭 is equal to x^{(q−1)/n}; the multiplicative character χ is defined on F_{q}^{∗} by χ(x) = ϵ, with χ(0) = 0. This character extends to a character on 𝔬 which we also denote by χ. Since χ is the

standard n^{th} power residue symbol, we can write it as χ(x, 𝔭) = (\begin{matrix}x \\ 𝔭 \end{matrix})_{n}; it is

its dependence on 𝔭 rather than on x which will mainly concern us.

Choose also a nontrivial additive character ψ on F_{q}; then the Gauss sum g(r) for r a non-zero element of Z/n is defined as

g(r) = − ∑_{x∈F_{q}^{∗}} χ^{r}(x)ψ(x).

Any other nontrivial additive character ψ′ has the form ψ′(x) = ψ(xξ) for some ξ in F_{q}^{∗}; and replacing ψ by ψ′ has the effect of replacing g(r) by

g′(r) = − ∑ χ^{r}(x)ψ(xξ) = − ∑ χ^{r}(xξ^{−1})ψ(x) = χ^{−r}(ξ)g(r). (2)

Replacing 𝔭 by 𝔭̄ takes χ into χ̄ and therefore replaces g(r) by g(−r). Now

g(r)g(−r) = ∑ ∑ χ^{r}(xy)ψ(xy)χ^{−r}(x)ψ(x) = ∑ ∑ χ^{r}(y)ψ(x(y + 1)).

Here the sum over x is −1 unless y = −1, so g(r)g(−r) = χ^{r}(−1)q; and similarly |g(r)|^{2} = q.

For any m > 1 we define the Jacobi sum to be

J(r, 𝔭) = J(r_{0}, …, r_{m}; 𝔭) = q^{−1}g(r_{0}) ⋯ g(r_{m})

where the r_{i} satisfy

r_{0} + … + r_{m} ≡ 0 mod n, n |̸r_{i}.

In particular J(r, 𝔭)J(r, 𝔭̄) = q^{m−1}. (In these definitions we follow [8]. The reader is warned that conventions for the Gauss and Jacobi sums vary from one author to another; in particular the three relevent papers by Weil employ three different conventions.)

Lemma 1 There are {(n − 1)^{m+1} − (−1)^{m}(n − 1)}/n such sets (r_{0}, …, r_{m}).

Proof Suppose there are A(m) such sets; by counting the number of solutions of ∑ r_{i} ≡ 0 mod n with 0 ≤ r_{i} < n in two different ways we obtain

n^{m} = ∑_{μ=0}^{m} (\begin{matrix}m + 1 \\ μ \end{matrix}) A(m − μ) + 1

where the bracket is a binomial coefficient. By induction on m, A(m) is a polynomial in n of degree at most m. It is therefore enough to find a formula for A(m) under the additional hypothesis that n is prime.

For any c in Z and fixed prime n let N(c) be the number of sets with

r_{0} + … + r_{m} ≡ c mod n, 0 < r_{i} < n.

If w is prime to n there is a one-one correspondence between the sets counted in N(c) and those counted in N(cw), obtained by multiplying each r_{i} by w and reducing mod n. Hence N(1) = … = N(n − 1) and therefore

N(0) + (n − 1)N(1) = (n − 1)^{m+1}.

Again, writing ζ = ζ_{n},

(−1)^{m+1} = (ζ + … + ζ^{n−1})^{m+1} = ∑ ζ^{c}N(c) = N(0) − N(1).

Since N(0) = A(m), the Lemma follows from these two equations. □

J(r, 𝔭) is independent of the choice of ψ, and Gal(k(\sqrt[p]{1})/k) acts on g(r) through its action on ψ. Hence by (2) it acts trivially on J(r, 𝔭), which must therefore be in k; similarly (g(r))^{n} must be in k.

Everything involving the r_{i} only depends on them mod n; but in order to state Stickelberger’s theorem we define θ(r) for any r in Q which is integral at n to be that integer which satisfies

0 ≤ θ(r) < n, θ(r) ≡ r mod n.

Weil’s version of Stickelberger’s theorem [7] shows that as an ideal (g(r))^{n} is principal with prime ideal decomposition

(g(r))^{n} = ∏(σ_{−c}^{−1}𝔭)^{θ(rc)}

where σ_{u} is the automorphism ζ ↦ ζ^{u} of k and the product is taken as c runs through a set of representatives of the classes in (Z/n)^{∗}. Throughout this paper, all products over c not otherwise specified will take this form. We can view σ_{u} as an element of Gal(k(\sqrt[p]{1})/Q(\sqrt[p]{1})), so that σ_{u}g(r) = g(ru). In particular σ_{−1} is complex conjugacy. As ideals

(J(r_{0}, …, r_{m}); 𝔭)^{n} = q^{−n} ∏(σ_{−c}^{−1}𝔭)^{θ(r_{0}c)+...+θ(r_{m}c)}.  (3)

Here θ(r_{0}c) + … + θ(r_{m}c) ≡ r_{0}c + … + r_{m}c ≡ 0 mod n, so that the right hand side is the n^{th} power of an ideal. Write for convenience

θ(r_{0}) + … + θ(r_{m}) = n{1 + α(r_{0}, …, r_{m})} = n{1 + α(r)}

where α is integral with 0 ≤ α < m; thus α(rc) + α(r(n − c)) = m − 1. Taking n^{th} roots in (3) gives

(J(r_{0}, …, r_{m}); 𝔭) = ∏(σ_{−c}𝔭)^{α(rc^{−1})}  (4)

as ideals. In particular, the right hand side of (4) is a principal ideal, which we write as (κ). Since every conjugate of J has absolute value q^{(m−1)/2}, it is possible to choose κ to have this same property, and now κ is determined up to a root of unity lying in k. To identify the value of J it is therefore enough to find J mod (1 − ρ)^{2}. If k has class number 1 we can write 𝔭 = (π) and κ = ∏(σ_{−c}π)^{α(rc^{−1})}. Here 𝔭 does not determine κ uniquely, because κ is changed if we multiply π by a power of ζ. But if also n = P or 2P, by multiplication by a suitable power of ρ we can require π to be congruent to a rational integer mod (1 − ρ)^{2}. Although π is still only determined up to a totally real unit, κ is then determined uniquely. Moreover, if π ≡ b mod (1 − ρ) then

κ ≡ ∏ b^{α(rc^{−1})} ≡ b^{m(P −1)/2} ≡ ±1 mod (1 − ρ),

so by modifying the previous recipe we can require κ ≡ 1 mod (1 − ρ)^{2}.

Write a^{r} = a_{0}^{r_{0}} ⋯ a_{m}^{r_{m}}. Weil [5] shows that the roots of the L-function of W̃ over F_{q} are the various χ̄(a^{r})J(r, 𝔭) where the r_{i} satisfy 0 < r_{i} < n and r_{0} + … + r_{m} ≡ 0 mod n, together with q^{(m−1)/2} if m is odd.

We can write the L-function of W as the product of simpler L-functions. Apart from ζ_{k}(s − (m − 1)/2) when m is odd, each of them has the form

M(s) = ∏_{𝔭}′(1 − J(r, 𝔭)χ̄(a^{r})(Norm 𝔭)^{−s})^{−1}.

(We use the letter M instead of L to indicate that we have left out possible factors corresponding to the bad primes.) For such an L-function to be geometric it is necessary that m is odd and α(cr) = \frac{1}{2}(m−1) for all c, so that each characteristic root at 𝔭 is q^{(m−1)/2} times a root of unity; conjecturally this is also sufficient. One case when this condition holds is when, after permutation of the r_{i} if necessary, r_{0} + r_{1} = r_{2} + r_{3} = … = n. For surfaces with n < 1000 this is the only case, and I believe this to be so in general. This is closely linked to the assertion (which has been proved in some cases; see for example Shioda [4]) that in middle dimension the geometric part of the cohomology is generated by the obvious linear subspaces.

3. Evaluation of certain Jacobi sums For any ideal 𝔟 in 𝔬 prime to n and the a_{i} we can now define J(r, 𝔟) by multiplicativity. Weil shows that J(r, 𝔟) considered as a function of 𝔟 is a Grossencharacter and n^{2} is a defining ideal for it. For fixed r and n a formula for J(r, 𝔟) can therefore be obtained by a finite calculation; and such formulae for small values of n are given in Berndt et al [1]. Tate has obtained a general formula for J(r, 𝔟), but for present purposes the following simpler results will suffice.

Lemma 2 Suppose that n = P or n = 2P where P is an odd prime, and that q = p^{ν} with ν even. Then J(r, 𝔭) = (−1)^{m+1}q^{(m−1)/2} and χ(a^{r}) = 1.

Proof Since ν is the smallest positive value of s for which p^{s} ≡ 1 mod n, we must have q^{1/2} = p^{ν/2} ≡ −1 mod n. We have ψ(x) = η^{Tr(x)} for some nontrivial p^{th} root of unity η, where the trace is for F_{q}/F_{p} and F_{p} is identified with Z/(p). Since Gal(F_{q}/F_{p}) is generated by x ↦ x^{p}, for any r

g(r) = − ∑ χ(x^{pr})ψ(x^{p}) = − ∑ χ(x^{pr})ψ(x) = g(rp).

In particular g(r) = g(rp^{ν/2}) = g(−r), and we know that g(r)g(−r) = χ^{r}(−1)q. If n = P is an odd prime, then

g(r) ≡ − ∑_{x≠0} ψ(x) ≡ 1 mod (1 − ρ),

so we must have χ^{r}(−1) = 1 and g(r) = −q^{1/2}. If n = 2P then, mod (1 − ρ),

g(r) ≡ − ∑_{x≠0}(χ(x))^{P r}ψ(x) ≡ \begin{cases}g(P r) & if r is odd, \\ 1 & if r is even. \end{cases}

If r is odd and ψ(x) = η^{Tr(x)} then −g(P r) is a quadratic Gauss sum, which is equal to −(−δ)^{ν}p^{ν/2} where δ = 1 if p ≡ 1 mod 4 and δ = i if p ≡ 3 mod 4. Thus g(r) is congruent to −1 mod (1 − ρ) if r is odd and p ≡ 3 mod 4 and 2||ν, and to 1 otherwise. So g(r) is q^{1/2} in the former case and −q^{1/2} in the latter. Since the former case happens an even number of times in the product for J, this proves the formula for J.

Since p − 1 is prime to P, every element of F_{p} is a P^{th} power; and since ν is even, every element of F_{p} is a square in F_{q}. So χ(a^{r}) = 1. □

Lemma 3 If n = P is an odd prime then J(r) ≡ 1 mod (1 − ρ)^{2}.

Proof For any x in F_{q}^{∗} define h(x) in Z/(n) by χ(x) = ζ^{h(x)}. For this Lemma ζ = ρ and so

(χ(x))^{r} ≡ 1 − rh(x)(1 − ρ) mod (1 − ρ)^{2};

therefore

g(r) ≡ 1 + r(1 − ρ) ∑ h(x)ψ(x) mod (1 − ρ)^{2}.

Thus J(r) ≡ 1 mod (1 − ρ)^{2}. □

Lemma 4 Suppose that r_{0}, r_{1} are integers not divisible by n. Then

g(r_{0})g(r_{1}) = \begin{cases}(χ(−1))^{r_{0}}q & if n|(r_{0} + r_{1}), \\ g(r_{0} + r_{1})J^{∗}(r_{0}, r_{1}) & otherwise, \end{cases}

where J^{∗}(r_{0}, r_{1}) = − ∑ χ(x^{r_{0}}(1 − x)^{r_{1}}).

Proof When n|(r_{0} +r_{1}) the result has already been proved; so we can assume that n |̸(r_{0} + r_{1}). Now

g(r_{0})g(r_{1}) = ∑ ∑(χ(xy))^{r_{0}}(χ(x))^{r_{1}}ψ(x(1 + y))

= ∑(χ(−1))^{r_{0}}(χ(x))^{r_{0}+r_{1}} + ∑ ∑_{y≠−1} (χ (\frac{xy}{1 + y} ))^{r_{0}} (χ (\frac{x}{1 + y} ))^{r_{1}} ψ(x)

= −g(r_{0} + r_{1}) ∑_{y≠−1} (χ (\frac{y}{1 + y} ))^{r_{0}} (χ (\frac{1}{1 + y} ))^{r_{1}} = g(r_{0} + r_{1})J^{∗}(r_{0}, r_{1}).

This completes the proof of the Lemma. □ It follows from (4) that as ideals

(J^{∗}(r_{0}, r_{1})) = ∏_{c}(σ_{−c}^{−1}𝔭)^{α(rc)}

where r = {r_{0}, r_{1}, −r_{0} − r_{1}}. Moreover J^{∗}(r_{0}, r_{1}) can be extended from 𝔭 to 𝔟 by multiplicativity, and as a function of 𝔟 it is a Grossencharacter and n^{2} is a defining ideal for it; this follows by Lemma 4 from the corresponding result for J, since J^{∗}(r_{0}, r_{1}) = (χ(−1))^{r_{0}+r_{1}}J(r_{0}, r_{1}, −r_{0} − r_{1}).

Lemma 5 Let n = 2P where P is an odd prime. Then, mod (1 − ρ)^{2}

J^{∗}(r_{0}, r_{1}) ≡ \begin{cases}1 & if r_{0}, r_{1} are both even, \\ (χ(2))^{2r_{1}} & if r_{0} is odd and r_{1} even, \\ χ(−1)(χ(2))^{−2(r_{0}+r_{1})} & if r_{0}, r_{1} are both odd. \end{cases}

Proof As in the proof of Lemma 3, write χ(x) = ζ^{h(x)}. Suppose first that r_{1} is even. Let ρ_{0} = 0 if r_{0} is even but ρ_{0} = P if r_{0} is odd. If x ≠ 0 we have

(χ(1 − x))^{r_{1}} ≡ \{1 − \frac{1}{2}r_{1}(1 − ρ)h(1 − x)\} mod (1 − ρ)^{2}

and (χ(1 − x))^{r_{1}} ≡ 1 mod (1 − ρ). Since (χ(x))^{r_{0}} ≡ (χ(x))^{ρ_{0}} mod (1 − ρ),

J^{∗}(r_{0}, r_{1}) ≡ 1 + \frac{1}{2}r_{1}(1 − ρ) ∑_{x≠0,1}(χ(x))^{ρ_{0}}h(1 − x) mod (1 − ρ)^{2}.

In particular, if r_{0} is also even then J^{∗}(r_{0}, r_{1}) ≡ 1 mod (1 − ρ)^{2}. For fixed y in F_{q} the number of solutions of x(1 − x) = y is 1 + χ^{P} (1 − 4y); so

\begin{aligned}J^{∗}(r, r) & = − ∑(χ(x(1 − x)))^{r} = − ∑\{1 + χ^{P} (1 − 4y)\}(χ(y))^{r} \\ & = − ∑(χ(4y))^{r}χ^{P} (1 − 4y)/χ(4^{r}) = J^{∗}(r, P)/χ(4^{r}). \end{aligned}

Taking r = 2 gives J^{∗}(2, P) ≡ χ(2^{4}) mod (1 − ρ)^{2} and therefore

∑(χ(x))^{P} h(1 − x) ≡ −(1 − χ(2^{4}))/(1 − ρ) mod (1 − ρ).

Hence if r_{1} is even and r_{0} is odd then

J^{∗}(r_{0}, r_{1}) ≡ 1 − \frac{1}{2}r_{1}(1 − χ(2^{4})) mod (1 − ρ)^{2}

and therefore

J^{∗}(r_{0}, r_{1}) ≡ (χ(2))^{2r_{1}} mod (1 − ρ)^{2}.

If r_{0}, r_{1} are both odd then

\begin{matrix}J^{∗}(r_{0}, & r_{1}) = − ∑_{x≠0,1}(χ(x^{−1})^{r_{0}}(χ(1 − x^{−1}))^{r_{1}} \\ = − ∑(χ(x))^{−r_{0}−r_{1}}(χ(x − 1))^{r_{1}} = χ(−1)J^{∗}(−r_{0} − r_{1}, r_{1}) \end{matrix}.

Using the previous result, this completes the proof of the Lemma. □

Corollary If n = 2P then J(r) ≡ (χ(−1))^{s/2}(χ(2))^{2R} mod (1 − ρ)^{2} where s is the number of odd r_{j} and R is the sum of the even r_{j}. If k has class number 1 we can choose κ ≡ 1 mod (1 − ρ)^{2} and then

J(r) = κ(χ(−1))^{s/2}(χ(2))^{2R}.

Proof When m = 1 this is the first case in Lemma 4. The general result then follows by induction on m, using Lemmas 4 and 5. □

Lemma 6 Suppose that n = 4 and p = q ≡ 1 mod 4. Let 𝔭 = (π) with π ≡ 1 mod 2(1 + i) be a prime factor of p in k = Q(i). Then

\begin{matrix}J^{∗}(1, 1) = χ(−1)π, J^{∗}(3, 3) = χ(−1)π̄, \\ J^{∗}(1, 2) = J^{∗}(2, 1) = π, J^{∗}(2, 3) = J^{∗}(3, 2) = π̄. \end{matrix}

Proof We have

J^{∗}(1, 2) = − ∑(1 − χ(x))(1 − χ^{2}(1 − x)) + p.

In this sum the terms with x = 0, 1 vanish and in every other term the first factor is divisible by 1+i and the second by 2. So J^{∗}(1, 2) ≡ 1 mod 2(1+i). Since (J^{∗}(1, 2)) = 𝔭 as ideals, J^{∗}(1, 2) = π.

For fixed y in F_{p} the number of solutions of x(1−x) = y is 1+χ^{2}(1−4y); so

\begin{aligned}J^{∗}(1, 1) & = − ∑ χ(x(1 − x)) = − ∑\{1 + χ^{2}(1 − 4y)\}χ(y) \\ & = − ∑ χ(4y)χ^{2}(1 − 4y)/χ(4) = J^{∗}(1, 2)/χ(4) \end{aligned}

and χ(−4) = 1. The remaining claims follow by complex conjugacy. □

Corollary Suppose that r contains n_{j} copies of j for j = 1, 2, 3, so that

n_{1} + n_{2} + n_{3} = m + 1, 4|(n_{1} + 2n_{2} + 3n_{3}).

Then

J(r) = (χ(−1))^{(n_{1}+n_{3})/2}π^{(3n_{1}+2n_{2}+n_{3})/4−1}π̄^{(n_{1}+2n_{2}+3n_{3})/4−1}.

Proof For J(1, 3) and J(2, 2) this is contained in Lemma 4. The general result now follows by induction on m, using Lemmas 4 and 6. □

4. The functional equation We now study the L-series

M(s) = ∏_{𝔭}(1 − J(r, 𝔭)χ̄(a^{r})(Norm 𝔭)^{−s})^{−1} = ∑_{𝔟}J(r, 𝔟) (a^{r} ; 𝔟̄)_{n} (Norm 𝔟)^{−s}.

Both the product and the series are absolutely convergent in ℜs > \frac{1}{2}(m + 1). When m = 4 or 6, we can analytically continue M(s) to ℜs > \frac{1}{2}(m − 1) in a way which, for certain integer values of s, expresses it as a finite sum in terms of Weierstrass elliptic functions; for m = 4 full details can be found below. But the corresponding process for other values of m introduces functions related to Abelian varieties of dimension \frac{1}{2}N with complex multiplication by 𝔬, where N is the order of Gal(k/Q) ≈ (Z/(n))^{∗}; and the theory of such functions is not yet explicit enough to meet our needs. We therefore have to proceed in a different way.

We begin by obtaining the functional equation, using the methods of Hecke rather than those of Tate’s thesis. One reason for this choice is that Hecke’s method gives more information about the theta series. Define Ψ by

J(r, (β)) (\begin{matrix}a^{r} \\ (β̄) \end{matrix})_{n} = Ψ(β) ∏_{c}(σ_{−c}β)^{α(rc^{−1})},

so that Ψ(β) as a function of a^{r} is a Dirichlet character whose values are powers of ζ and n^{2} ∏ a_{i} is a defining ideal for it. The only property of Ψ that we shall actually use is that Ψ(σβ) = σΨ(β) for all β in 𝔬 and σ in Gal(k/Q); this implies that the conductor of Ψ is invariant under Gal(k/Q), and in particular that it is principal. Let 𝔣 = (f) be the conductor of Ψ.

The units of k are generated by the totally real units and the roots of unity, together with 1−ζ_{n} if n is not a prime or twice an odd prime. (See [6], Corollary 4.13.) Hence Ψ is trivial on all totally real units and Ψ(ζ) = ζ^{A(r)} where A(r) = ∑_{c} θ(cα(rc^{−1})). When n is prime the case when A(r) ≡ 0 mod n is of some interest. This can happen even if m = 2. A systematic example is when n ≡ 1 mod 3 and r_{0}^{2} + r_{0}r_{1} + r_{1}^{2} ≡ 0 mod n; and there are also sporadic examples, the simplest being when n = 59 and r_{1} = 3r_{0}. ρ is a root of unity. If 1 − ζ_{n} is a unit then we also have Ψ(1 − ζ_{n}) = ζ_{n}^{A(r)/2}. (It can be shown directly that in this case A(r) is even whenever n is.)

Denote by C the set of odd integers c in Z prime to n and satisfying |c| < n if n is odd or |c| < \frac{1}{2}n if n is even, and by C^{+} the set of positive elements of C. For fixed r and 0 ≤ j < m write

C_{j} = \{c | c in C, α(c^{−1}r) = j\} for j ≠ \frac{1}{2}(m − 1);

thus if c is in C_{j} then −c is in C_{m−1−j}. If m is odd write also

C_{(m−1)/2} = \{c | c in C^{+}, α(c^{−1}r) = \frac{1}{2}(m − 1)\}.

Thus c in C_{j} implies that for any β in k^{∗}

(σ_{c}β)^{α(−c^{−1}r)}(σ_{−c}β)^{α(c^{−1}r)} = (σ_{c}β)^{m−1−2j}{(σ_{c}β)(σ_{−c}β)}^{j}.

To abbreviate some formuae, in what follows we shall write ∏^{∗} for

∏_{j=0}^{(m−1)/2} ∏_{c in C_{j}} .

We can now suppress the dependence on W_{m,n} and α. Instead we partition the elements of C^{+} among the sets C_{j} where 0 ≤ j < m. If c is in C_{j} and j ≠ \frac{1}{2}(m − 1) we then assign −c to C_{m−j−1}. This leaves in limbo the −c for which c is in C_{(m−1)/2}; they should be thought of as nearly but not quite in C_{(m−1)/2}. In this notation we have

\begin{aligned}L(s) & = ∑_{𝔟}Ψ(β)(Normβ)^{−s}∏^{∗}\{(σ_{−c}^{−1}β)^{j}(σ_{c}^{−1}β)^{m−1−j}\} \\ & = ∑_{𝔟}Ψ(β)∏^{∗}\{(σ_{c}^{−1}β)^{m−1−2j}(σ_{−c}^{−1}β.σ_{c}^{−1}β)^{j−s}\}. \end{aligned}

For simplicity we now assume that k has class number 1. (If this hypothesis is dropped, the argument which follows only needs some further notational complications; but I see no point in extending the computations to fields with class number greater than 1.) Write

ϑ^{0}(y, v, 𝔣) = ∑_{α∈𝔬}exp{−2π ∑_{C^{+}} y_{c}d_{c}(σ_{c}(fα) + v_{c})(σ_{−c}(fα) + v_{−c})}

where the y_{c} and d_{c} are real and positive. We shall choose the d_{c} later. The v_{c} will satisfy v_{c} = ∑ ζ_{n}^{ct}u_{t} where the u_{t} are real; thus we can identify the space of v with the space of u, which is R^{n}. In particular we shall use this identification to determine a measure on the v-space. Since ϑ^{0} is not affected by multiplying f by a unit, it does not depend on the choice of f. The series for ϑ^{0}, like all the analogous theta series which we shall consider, is absolutely convergent and uniformly so provided y and v lie in bounded domains and each y_{c} is bounded away from 0. This is enough to justify all the manipulations which follow.

Let Λ be the lattice of points u in Q^{N} for which the associated v are such that each v_{c} is in 𝔣. Thus v_{c} = σ_{c}v_{1} for such v and ϑ^{0} is periodic in u with Λ as its lattice of periods. Let 𝔡 = (δ) be the different of k; thus

𝔡 = (n ∏_{p|n}(1 − exp(2πi/p))^{−1}).

For μ in k with μ_{c} = σ_{c}μ, ∑ μ_{c}v_{c} is in Z for each v associated with some point of Λ if and only if μ is in (𝔡𝔣)^{−1}. Thus ϑ^{0} has a Fourier expansion

ϑ^{0} = ∑_{μ∈(𝔡𝔣)^{−1}} b(μ) exp{2πi ∑_{C} μ_{c}v_{c}}.

If V is the volume of R^{N}/Λ then

b(μ) = V^{−1} ∫ ϑ^{0} exp{−2πi ∑ μ_{c}v_{c}}du

= V^{−1} ∫ exp{−2π(∑_{C^{+}} y_{c}d_{c}v_{c}v_{−c} + i ∑_{C} μ_{c}v_{c})}du

where the first integral is taken over R^{N}/Λ and the second integral is taken over R^{N}. In the last integral we replace each v_{c} by v_{c} − iμ_{−c}/y_{|c|}d_{|c|}. Doing so shifts each interval (−∞, ∞) of integration by an imaginary constant, but by a standard argument we can shift it back; hence we obtain

b(μ) = V^{−1}I exp{−2π ∑_{C^{+}} μ_{c}μ_{−c}/y_{c}d_{c}}

where

I = ∫_{−∞}^{∞} … ∫_{−∞}^{∞} exp{−2π ∑_{C^{+}} y_{c}d_{c}v_{c}v_{−c}}du.

This integral can be evaluated explicitly; but all we need to know is that it is equal to a constant positive multiple of ∏(y_{c}d_{c})^{−1}. Thus

ϑ^{0} = V^{−1}I ∑_{μ} exp{−2π ∑_{C^{+}} μ_{c}μ_{−c}/y_{c}d_{c} + 2πi ∑_{C} μ_{c}v_{c}}.  (5)

This is an identity, subject only to the condition that the u_{c} are rational; and since both sides are analytic in the u_{c}, this condition can be dropped. If for each j < \frac{1}{2}m we differentiate (5) (m − 1 − 2j) times with respect to each v_{c} with c in C_{j} and write

\begin{matrix}B = & ∏^{∗}(y_{|c|}d_{|c|}(σ_{−c}(fα) + v_{−c}))^{m−2j−1}, \\ C = & ∏^{∗}(−iμ_{c})^{m−2j−1} \end{matrix}

we obtain

\begin{matrix}∑_{α} B exp\{−2π ∑_{C^{+}} y_{c}d_{c}(σ_{c}(fα) + v_{c})(σ_{−c}(fα) + v_{−c})\} \\ = V^{−1}I ∑_{μ} C [exp\{−2π ∑_{C^{+}} μ_{c}μ_{−c}/y_{c}d_{c} + 2πi ∑_{C} μ_{c}v_{c}\}] . \end{matrix}  (6)

Write β = \overline{μfδ}, so that β runs through the elements of 𝔬. On the right hand side of (6) the term in square brackets only depends on v mod {f/(β̄, f)} where the round bracket denotes the highest common factor. Now multiply (6) by Ψ(v) and sum over v as it runs through a set of representatives of the classes in 𝔬/𝔣. Since (f) is the exact conductor of Ψ, if β̄ is not prime to f we have ∑ Ψ(v) = 0; so the terms with β not prime to f contribute nothing to the sum on the right. We can therefore multiply each term in that sum by the corresponding Ψ(β)Ψ̄(β). Let d_{c} be such that d_{c}^{−2} = (σ_{c}(fδ))(σ_{−c}(fδ)); then on the right hand side we obtain

V^{−1}I ∑_{β} CΨ(β)W_{1}(Ψ) exp{−2π ∑_{C^{+}} d_{c}(σ_{c}β)(σ_{−c}β)/y_{c}}

where

W_{1}(Ψ) = ∑_{v} Ψ(β̄v) exp{2πiTr(β̄v/fδ)}

with v running through a set of representatives of the classes in (𝔬/𝔣)^{∗}. Since β is prime to 𝔣, there is one such β̄v in each class of (𝔬/𝔣)^{∗}; so we can write the last equation as

W_{1}(Ψ) = ∑_{v} Ψ(v) exp{2πiTr(v/fδ)}

where v runs through a set of representatives of (𝔬/𝔣)^{∗}. Here W_{1}(Ψ) does depend on the choice of δ and f. Explicitly, if we multiply δ or f by a unit ϵ we also multiply W_{1} by Ψ(ϵ). The function ϑ also depends on the choice of δ, but in a more complicated way.

On the left hand side we obtain

∑_{β} Ψ(β) exp{−2π ∑_{C^{+}} y_{c}d_{c}(σ_{c}β)(σ_{−c}β)}∏^{∗}(y_{|c|}d_{|c|}σ_{−c}β)^{m−2j−1}

where the sum is over all elements β = fα + v of 𝔬. Thus if we write

θ(y, Ψ) = ∑_{β} Ψ(β) exp{−2π ∑_{C^{+}} y_{c}d_{c}(σ_{c}β)(σ_{−c}β)}∏^{∗}(σ_{−c}β)^{m−2j−1}

then we have

θ(y, Ψ)∏^{∗}y_{|c|}^{m−2j} = W θ(y^{−1}, Ψ)  (7)

for some W independent of the y_{c}, where y^{−1} is the vector of the y_{c}^{−1} for c in C^{+}. In other words, replacing y by y^{−1} multiplies

θ(y, Ψ)∏^{∗}y_{|c|}^{m/2−j}

by W. It follows from this that W^{2} = 1. In fact we have

W = V^{−1}I (∏_{C^{+}} y_{c}) (∏^{∗}(iσ_{c}(fδ)d_{|c|})^{2j+1−m}) W_{1}(Ψ).

The Mellin transform formula gives

∫_{0}^{∞} exp\{−2πyd_{c}(σ_{c}β)(σ_{−c}β)\}y^{s−j} \frac{dy}{y} = Γ(s − j)\{2πd_{c}(σ_{c}β)(σ_{−c}β)\}^{j−s}

for c in C^{+}. Then provided ℜs > \frac{1}{2}(m + 1),

L(s)∏^{∗} \frac{Γ(s − j)}{(2πd_{|c|})^{s−j}} = ∫_{0}^{∞} ⋯ ∫_{0}^{∞} θ^{∗}(y, Ψ)∏^{∗} (y_{|c|}^{s−j} \frac{dy_{|c|}}{y_{|c|}} )  (8)

where

θ^{∗}(y, Ψ) = ∑_{𝔟}Ψ(β) exp{−2π ∑_{C^{+}} y_{c}d_{c}(σ_{c}β)(σ_{−c}β)}∏^{∗}(σ_{−c}β)^{m−2j−1}.

We now modify (8) in such a way as to replace θ^{∗} by θ — that is, to replace the sum over all 𝔟 by a sum over all β. Let the ξ^{(c)} for c in C^{+} be multiplicatively independent totally real units in k, let G be the group which they generate and G^{(2)} the group of γ^{2} with γ in G; and write g = [𝔬^{∗} : G^{(2)}]. Suppose that in θ^{∗} we replace β by ϵβ, where ϵ is a unit in 𝔬; then provided we divide each y_{c} by (σ_{c}ϵ)(σ_{−c}ϵ) the integrand in (8) is unchanged. This operation defines an action of 𝔬^{∗} on the space of points y. Thus we can replace the sum in θ^{∗} by a sum over a set of representatives of all the classes in {(𝔬 \ {0})/G^{(2)}} provided we also divide the right hand side of (8) by g. Let z^{N/2} = ∏ y_{c} and write η_{c} = y_{c}/z, so that the η_{c} are positive with product 1. Let V_{1} be a fundamental domain for the action of G^{(2)} on η chosen to be invariant under η ↦ η^{−1}; then the integral on the right hand side of (8) is just the sum of the integrals over (0, ∞)×γ^{2}V_{1} as γ runs through the elements of G. In each such integral we write γ_{c}^{2}y_{c} for y_{c}; if we also replace β by γ^{−1}β, this leaves the integrand unchanged. So (8) becomes

L(s)∏^{∗} \frac{Γ(s − j)}{(2πd_{|c|})^{s−j}} = g^{−1} ∫ θ(y, Ψ)∏^{∗} (y_{|c|}^{s−j} \frac{dy_{|c|}}{y_{|c|}} )

where the integral is taken over (0, ∞)×V_{1}. Now split the range of integration for z as the union of the two subintervals (0, 1) and (1, ∞) and use (7) on the former; then we obtain

\begin{matrix}L(s) & ∏^{∗} \frac{Γ(s − j)}{(2πd_{|c|})^{s−j}} = \\ g^{−1} ∫ \{θ(y, Ψ)∏^{∗}y_{|c|}^{s−j} \frac{dy_{|c|}}{y_{|c|}} + W θ(y, Ψ)∏^{∗}y_{|c|}^{m−s−j} \frac{dy_{|c|}}{y_{|c|}} \} , \end{matrix} (9)

where the integral is now taken over (1, ∞)×V_{1}. The right hand side provides the analytic continuation of L(s) to the whole s-plane, and we also obtain the functional equation. The latter says that

L(s)∏^{∗} \frac{Γ(s − j)}{(2πd_{|c|})^{s−j}}

is multiplied by W if we replace s by m − s.

However, the multiple integral which we have obtained for L(s) in (9) is not usually convenient for numerical calculation, and it is better to proceed in a less symmetric way. Let y′ in R^{N/2−1} be the point whose coordinates are the y_{c} with c in C^{+} other than y_{1} and let V_{2} ⊂ R^{N/2−1} be a fundamental domain for the action of G^{(2)} on y′ chosen to be invariant under y′ ↦ y′^{−1}. An argument like that which led to (9) leads to an equation which is typographically the same but in which the multiple integral is taken over y′ in V_{2} and 1 ≤ y_{1} ≤ ∞. Provided the value of s is in Z and C_{j} is empty whenever \frac{1}{2}(m − 1) ≥ j ≥ min(s, m − s), the multiple integral can then be evaluated explicitly. (Otherwise its value can be computed numerically, but this is no help towards identifying it.) In particular this can be done in the special case when m is even and s = \frac{1}{2}m. There are two reasons why this case is important. If the sign in the functional equation is −1 then L(\frac{1}{2}m) = 0. And if the sign is +1 then one hopes that L(\frac{1}{2}m) is the product of an identifiable transcendental, a fudge factor and an integral square.

5. The case n = 4. Now k = Q(i). Suppose that α, α′ are elements of 𝔬 which are odd and coprime. The laws of biquadratic reciprocity state that if α = λ + iμ and α′ = λ′ + iμ′ are both congruent to 1 mod(2 + 2i), which can always be achieved by multiplying α, α′ by suitable powers of i, then

(\begin{matrix}α \\ α′ \end{matrix})_{4} = (\begin{matrix}α′ \\ α \end{matrix})_{4} (−1)^{μμ′/4}, (\begin{matrix}1 + i \\ α \end{matrix})_{4} = i^{(λ−μ−1−μ^{2})/4}, (\begin{matrix}2 \\ α \end{matrix})_{4} = i^{−μ/2}.

To simplify the interpretation of the transcendental factor of L(s) for strategic s, it is advantageous to make use of Weierstrass functions. Let ω = 2 · 6220575543 … be such that the periods of the Weierstrass function satisfying ℘′^{2} = 4℘^{3} − 4℘ are ω, iω. The most convenient way to evaluate ℘(z) is by means of the formula

\frac{ω^{2}℘(z)}{(2πi)^{2}} = \frac{Q}{(1 − Q)^{2}} + ∑_{r=1}^{∞} \{\frac{q^{r}Q}{(1 − q^{r}Q)^{2}} + \frac{q^{r}Q^{−1}}{(1 − q^{r}Q^{−1})^{2}} \} + \frac{1}{4π}

where q = exp(−2π) and Q = exp(2πiz/ω). The corresponding formulae for ℘′(z) and for the Weierstrass zeta function are

\frac{ω^{3}℘′(z)}{(2πi)^{3}} = \frac{Q(1 + Q)}{(1 − Q)^{3}} + ∑_{r=1}^{∞} \{\frac{q^{r}Q(1 + q^{r}Q)}{(1 − q^{r}Q)^{3}} − \frac{q^{r}Q^{−1}(1 + q^{r}Q^{−1})}{(1 − q^{r}Q^{−1})^{3}}\} ,

\frac{ωζ(z)}{2πi} = \frac{1}{2} − \frac{1}{1 − Q} − \frac{iz}{2ω} + ∑_{r=1}^{∞} \{−\frac{q^{r}Q}{1 − q^{r}Q} + \frac{q^{r}Q^{−1}}{1 − q^{r}Q^{−1}} \} .

Moreover

ζ(z + ω) = ζ(z) + π/ω, ζ(z + iω) = ζ(z) − iπ/ω.

The case m = 2 is already in [2]. The only kind of admissible triad is (u, u, 2u), so that we must consider

L^{a}(s) = ∑_{𝔞}χ(𝔞)ᾱ(αᾱ)^{−s}  (10)

where 𝔞 = (α) with α ≡ 1 mod (2 + 2i). If a > 1 is a positive integer which is fourth power free, it is shown in [2] that

a^{1/4}L^{a}(1)/ω and (4a)^{1/4}L^{−a}(1)/ω

are rational integers, and their values are extensively tabulated there. The relevent part of the Birch/Swinnerton-Dyer conjecture is as follows:

L^{a}(1) = (ωa^{−1/4})F X^{2}

where X^{2} is the order of the Tate-Safarevic group and F = F^{∗} ∏ F_{p} where the product is taken over the odd primes dividing a. For such p F_{p} = 2, which is the Tamagawa number of the curve y^{2} = 4x^{3} − 4ax, and F^{∗} is \frac{1}{16} if a is a square and \frac{1}{4} otherwise. (Thus F is almost but not quite a product of local factors.) Recall also that the Tate-Safarevic group is an H^{1}.

The case m = 3 is the one considered in [3]; we summarize here the relevent part of the conclusions. The two tetrads which give rise to transcendental L-functions are (1, 1, 1, 1) and (3, 3, 3, 3), and it is enough to consider the first of these. Denote by a the fourth-power-free part of a_{0}a_{1}a_{2}a_{3} and

write χ(𝔞) = (\begin{matrix}a \\ 𝔞 \end{matrix})_{4}; up to a possible factor associated with the prime 2 the

L-series is

M(a, s) = ∏′ᾱ^{2}(αᾱ)^{−s}χ(𝔞)

where the sum is taken over all odd ideals 𝔞 in Z[i] prime to a and 𝔞 = (α) with α ≡ 1 mod (2 + 2i). Let

a = 2^{v}ϵc_{0} with ϵ = ±1, c_{0} ≡ 1 mod 4.  (11)

Because −4 is a fourth power in Q(i), we need only consider a > 0. Let ψ be the character satisfying ψ(α) = χ(𝔞) where 𝔞 = (α) and α ≡ 1 mod (2+2i); and extend this by writing ψ(i) = −1. The odd part of 𝔣, the conductor of ψ, is the product of the odd primes dividing a and the even part is

8 if v is odd,
4 if v ≡ 1 + ϵ mod 4,
2 if v ≡ 1 − ϵ mod 4 and c_{0} ≡ 1 mod 8,
1 if v ≡ 1 − ϵ mod 4 and c_{0} ≡ 5 mod 8.

Choose f > 0 so that 𝔣 = (f). We set ψ(1 + i) = i^{(1−c_{0}^{2})/8} if f is odd; the missing factor in the L-series is then

{1 − (−1)^{(c_{0}−5)/8}2^{1−s}}^{−1}.

With this factor included, we denote the L-series by L(a, s). Then

L(a, 2) = \frac{1}{4}ω^{2}f^{−2} ∑ ψ(u)℘(ωu/f)

where the sum is taken over a set of representatives u of (Z[i]/(f))^{∗}. The main conclusion of [3] is as follows:

Theorem 1 Suppose that a > 0 and f > 3; then a^{1/4}ff_{2}ω^{−2}L(a, 2) is in Z, where f_{2} is the even part of f.

The proviso f > 3 is essential here. The substantial tables in [3] suggest that, provided a > 0 and f > 3,

2^{−t}a^{1/4}fω^{−2}L(a, 2)  (12)

is a positive integer, where t is the number of odd prime factors of f. It is tempting to regard ω^{2}/a^{1/4}f as a scaling factor, to call 2^{t} a fudge factor and to conjecture that the integer (12) is the order of the H^{1} identified by Bloch and Kato. But in the three cases a = 1, 12 and 108 for which f < 4, (12) is not an integer, so the situation cannot be quite so simple.

As m increases, new L-functions appear. By the Corollary to Lemma 6 they all have the form ∑ χ(𝔞)α^{λ_{1}}ᾱ^{λ_{2}}(αᾱ)^{−s} where λ_{1}, λ_{2} are non-negative integers with λ_{1}+λ_{2} = m−1. By an obvious change of variables, it is enough to consider ∑ χ(𝔞)ᾱ^{μ}(αᾱ)^{−s} for some a, where m + μ is an odd integer. The more interesting case, which is the only one which will be treated here, is when m is even; the case when m is odd is similar except that the rule for the conductor is more complicated and it may be necessary to supply a factor for L(s) corresponding to the prime 2. Thus we write

L_{ν}(s) = ∑_{𝔞}χ(𝔞)ᾱ^{2ν+1}(αᾱ)^{−s}  (13)

where ν ≥ 0 is an integer and the sum is over all odd 𝔞 prime to a. (We drop the previous meaning of ν.) As usual we take ψ(α) = χ(𝔞) if 𝔞 = (α) and α ≡ 1 mod (2 + 2i), but this time we extend this definition by ψ(i) = i^{2ν+1}.

We again write a in the form (11). The odd part of 𝔣, the conductor of ψ, is the product of the odd primes dividing a; and this time the even part is

8 if v is odd,
4 if v ≡ 1 + ϵ mod 4,
2(1 + i) if v ≡ 1 − ϵ mod 4.

The strategic values of s are the integers from ν + 1 to 2ν + 1 inclusive. The most interesting of these is ν + 1, which is at the centre of the critical strip. To continue L_{ν}(s) into the critical strip we write

\begin{matrix}g_{ν}(z, s) = & z̄^{2ν+1}|z|^{−2s} + ∑_{u}′\{(z̄ + ū)^{2ν+1}|z + u|^{−2s} \\ − ū^{2ν+1}|u|^{−2s} − (2ν + 1 − s)z̄ū^{2ν}|u|^{−2s} + szū^{2ν+2}|u|^{−2−2s}\} \end{matrix}

where the sum is taken over all nonzero u in the lattice generated by ω and iω. This series is absolutely convergent in ℜs > ν + \frac{1}{2}.

In ℜs > ν + \frac{3}{2} we can split this last sum into four sums, each of which is absolutely convergent. Of these, ∑′ ū^{2ν+1}|u|^{−2s} vanishes identically because the terms cancel in pairs; and for each ν the same is true of one of the last two sums. Hence we have

\begin{matrix}g_{ν}(z + ω, s) − g_{ν}(z, s) = sωh_{ν+1}(s + 1) − (2ν + 1 − s)ωh_{ν}(s), \\ g_{ν}(z + iω, s) − g_{ν}(z, s) = siωh_{ν+1}(s + 1) + (2ν + 1 − s)iωh_{ν}(s) \end{matrix}

where we have written h_{μ}(s) = ∑′ ū^{2μ}|u|^{−2s}. In particular this implies that h_{ν}(s) and h_{ν+1}(s + 1) can be analytically continued to ℜs > ν + \frac{1}{2}. If μ is odd, then trivially h_{μ}(s) = 0. Now write

g_{ν}^{∗}(z, s) = g_{ν}(z, s) + (2ν + 1 − s)z̄h_{ν}(s) − szh_{ν+1}(s + 1),

so that g_{ν}^{∗}(z, s) is a doubly periodic function of z, though it is not analytic. Setting s = ν + 1 we have

g_{ν}^{∗}(z, ν + 1) = ∑_{j=0}^{ν} \frac{Γ(ν + 1)}{Γ(j + 1)Γ(ν − j + 1)}z̄^{j}g_{ν}^{j}(z)

where if ν > 1 g_{ν}^{j}(z) is equal to

∑ ū^{ν−j}(z + u)^{−ν−1} if j > 1,

∑′ū^{ν−1}{(z + u)^{−ν−1} − u^{−ν−1}} + h_{ν}(ν + 1) if j = 1,

∑′ū^{ν}{(z + u)^{−ν−1} − u^{−ν−1} + (ν + 1)zu^{−ν−2}} − (ν + 1)zh_{ν+1}(ν + 2) if j = 0.

If ν = 1 the first two lines should be replaced by g_{1}^{1} = ℘(z). Moreover g_{ν}^{j}(z) is an even function of z if j is odd and an odd function if j is even. Clearly

g_{ν}^{ν}(z) = (−1)^{ν−1}℘^{(ν−1)}(z)/Γ(ν + 1);

the other g_{ν}^{j}(z) are meromorphic but not doubly periodic, and near z = 0

g_{ν}^{j}(z) = h_{ν}(ν + 1) + O(z) for 0 < j < ν, g_{ν}^{0} = O(z).

To exploit the double periodicity we write

g_{ν}^{∗}(z, ν + 1) = ∑_{j=0}^{ν} \frac{Γ(ν + 1)}{Γ(j + 1)Γ(ν − j + 1)}(z̄ − π^{−1}ω^{2}ζ(z))^{j}f_{ν}^{j}(z)

where

f_{ν}^{j}(z) = ∑_{k=j}^{ν} \frac{Γ(ν − j + 1)}{Γ(k − j + 1)Γ(ν − k + 1)}(π^{−1}ω^{2}ζ(z))^{k−j}g_{ν}^{k}(z)

is meromorphic and doubly periodic. Its only possible poles are at the lattice points and its behaviour near z = 0 can be read off from that of g_{ν}^{j}(z). Once we have determined the f_{ν}^{j}(z) for j = ν, …, μ+1 this enables us to determine f_{ν}^{μ} up to the temporarily unknown constant C = h_{ν−(μ−1)/2}(ν + 1) when μ ≡ 2ν + 1 mod 4. This constant must be carried through to the next step, at which we find that the expression for f_{ν}^{μ−1}(z) contains a multiple of Cζ(z). The double periodicity of f_{ν}^{μ−1}(z) now determines C. When z is close to a lattice point, care must be taken in evaluating g^{∗} numerically, because the components of it are large compared to their sum.

Let f_{∗} be in Q ∩ 𝔣 and not divisible by any prime in 𝔬 which does not divide 𝔣. In (13) we write α = f_{∗}uω^{−1} + γ where γ ≡ 1 mod (2 + 2i) runs through a set of representatives of the relevent classes in (𝔬/f_{∗})^{∗}. Thus

L_{ν}(s) = (\frac{ω}{f_{∗}} )^{2s−2ν−1} ∑_{γ} χ(γ)g_{ν}^{∗}(γω/f_{∗}, s)  (14)

in ℜs > ν + \frac{3}{2}. Both sides are holomorphic in ℜs > ν + \frac{1}{2} so the conclusion extends to that region.

Similar arguments give closed formulae for L_{ν}(s) for s = ν +2, …, 2ν +1; the calculations are now simpler since we do not need to introduce g_{ν}(z, s). In particular

L_{ν}(2ν + 1) = \frac{−ω^{2ν+1}}{f_{∗}^{2ν+1}Γ(2ν + 1)} ∑_{γ} ψ(γ)℘^{(ν)}(γω/f_{∗}).

We illustrate this process by calculating L_{1}(2) and L_{2}(3). For L_{1}(2) we have h_{1}(s) = 0 and g_{1}^{1}(z) = ℘(z), whence

g_{1}^{∗}(z, 2) = (z̄ − π^{−1}ω^{2}ζ(z))℘(z) + f_{1}^{0}(z).

But f_{1}^{0}(z) = π^{−1}ω^{2}z^{−3} + O(z) near z = 0, so f_{1}^{0}(z) = −\frac{1}{2}π^{−1}ω^{2}℘′(z). Thus

g_{1}^{∗}(z, 2) = (z̄ − π^{−1}ω^{2}ζ(z))℘(z) − \frac{1}{2}π^{−1}ω^{2}℘′(z).

Such formulae are convenient for computing, but not for proving results about the nature of L_{ν}(s). For the latter purpose, we recall that

bζ(z) = \frac{bπz̄}{ω^{2}} − ∑_{r=1}^{b−2} \frac{℘′(rz) − ℘′(z)}{2(℘(rz) − ℘(z))}

for any b > 0 in Z, where zb/ω is an integer in Q(i) and the summand with r = 1 is

℘′′(z)/2℘′(z) = (3(℘(z))^{2} − 1)/℘′(z).

Using a sufficiently strengthened version of the Kronecker Jugendtraum, it ought to be possible to show that if s is a strategic integer

π^{2ν+1−s}ω^{−2ν−1}a^{1/4}f_{odd}^{s−1}L_{ν}(s)

is a rational integer divided by a power of 2. Here f_{odd} is the odd part of f — that is to say, the product of the odd primes which divide a; and we introduce it rather than the more natural |f| to prevent factors \sqrt{2} appearing in the following tables. Numerical evidence overwhelmingly suggests that when s = ν + 1 the underlying rational integer is a perfect square. Part of this evidence is the following table, which describes L_{1}(2) for all a divisible only by 2,3,5 and 7.

We expect to have

L_{1}(2) = (ω^{3}π^{−1}a^{−1/4}f_{odd}^{−1})F X^{2}

where the first factor is more or less the Bloch-Kato transcendental, F is a fudge factor and X^{2} is an integral square. (The factor f_{odd}^{−1} may really belong to F, but I prefer to believe that F is simply a power of 2.) It appears also that F = ∏ F_{p} where F_{p} is a local factor for the bad prime p; the numerical evidence does not suggest a need for a non-local factor of the kind that appears in the Birch/Swinnerton-Dyer conjecture. The numerical evidence can only determine the F_{p} up to powers of 4 (and evn then there is more than one possible pattern); we have chosen the F_{p} to make the X^{2} as small as possible. The table of values of F_{p} for odd p is as follows:

\begin{matrix}n_{p} odd & n_{p} odd & n_{p} = 2 & n_{p} = 2 \\ χ_{p} = 1 & χ_{p} = −1 & χ_{p} = 1 & χ_{p} = −1 \\ 2 & 2 & 4 & 2 \end{matrix}

The table of values of F_{2} is as follows, where the congruences for c_{2} are mod 16:

n_{2} = 0 n_{2} = 1 n_{2} = 2 n_{2} = 3
c_{2} ≡ 1  2  4  8  4
c_{2} ≡ 3  16  4  8  4
c_{2} ≡ 5  8  4  16  4
c_{2} ≡ 7  16  4  8  4
c_{2} ≡ 9  2  4  8  4
c_{2} ≡ 11 16  4  16  4
c_{2} ≡ 13 8  4  16  4
c_{2} ≡ 15 16  4  16  4

It is natural to hope that the F_{p} are related to Tamagawa numbers.

Question 3 How does one define and how does one computs the Tamagawa number of a motive? Are the values of F_{p} above equal to the Tamagawa numbers of the corresponding motives?

The table which follows gives the values of

X^{2} = πω^{−3}a^{1/4}f_{odd}(∏ F_{p})^{−1}L_{1}(2).

The first column gives the odd part of a. The second, third and fourth give respectively the exponents of 3,5 and 7 in a. The remaining four columns relate to the cases when the power of 2 in a is 1,2,4 or 8 respectively; the entries give the values of X^{2}.

1 0 0 0 1 0  0  1
3 1 0 0 0 0  1  1
9 2 0 0 1 1  1  0
27 3 0 0 0 0  0  9
5 0 1 0 1 0  1  1
25 0 2 0 0 0  0  1
125 0 3 0 0 0  1  9
7 0 0 1 1 0  1 16
49 0 0 2 0 1  4  0
343 0 0 3 1 0  9  4
15 1 1 0 1 0  0  9
75 1 2 0 0 0  0  4
375 1 3 0 4 0  1 25
45 2 1 0 1 4  0  0
225 2 2 0 0 25  1  0
1125 2 3 0 0 1  0  0
135 3 1 0 0 0  9  9
675 3 2 0 0 0  9  9
3375 3 3 0 9 0  0  9
21 1 0 1 9 0  1  1
147 1 0 2 1 25  0  0
1029 1 0 3 1 0 25  1
63 2 0 1 0 25  1  0
441 2 0 2 0 0  0  1
3087 2 0 3 0 9  4  0
189 3 0 1 0 0  9  9
1323 3 0 2 9 81  9  0
9261 3 0 3 0 0  9 81
35 0 1 1 0 0  1 25
245 0 1 2 0 49 0 0
1715 0 1 3 0 0 25  9

175 0 2 1 4  0  0  1
1225 0 2 2 1  16  9  0
8575 0 2 3 4  0  0  81
875 0 3 1 0  0  0  1
6125 0 3 2 9  25  0  0
42875 0 3 3 0  0  0  81
105 1 1 1 0  0  0  81
735 1 1 2 0  4  9  0
5145 1 1 3 0  0  0  25
525 1 2 1 0  0  25  81
3675 1 2 2 9  25  25  0
25725 1 2 3 0  0  4  361
2625 1 3 1 100  0  0  9
18375 1 3 2 0  4  0  0
128625 1 3 3 676  0  0  169
315 2 1 1 9  4  25  0
2205 2 1 2 0  0  4  1
15435 2 1 3 25  169  1  0
1575 2 2 1 0  100  0  0
11025 2 2 2 25  0  0  81
77175 2 2 3 0  121  0  0
7875 2 3 1 1  100  0  0
55125 2 3 2 64  0  100  169
385875 2 3 3 1  49  0  0
945 3 1 1 36  0  0  225
6615 3 1 2 0  9  0  0
46305 3 1 3 36  0  0  9
4725 3 2 1 9  0  0  81
33075 3 2 2 9  36  0  0
231525 3 2 3 144  0  9  441
23625 3 3 1 0  0  0  9
165375 3 3 2 0  225  36  0
1157625 3 3 3 0  0  0  1089

Write σ = ±1 for the sign in the functional equation. If the sign is −1 then L_{1}(2) = 0; and presumably the converse is usually true because one expects that L_{1}(s) only rarely has a zero of order greater than 1 at s = 2. (The only examples of such a zero in this table appear to be when a = 3^{3}.5 or a = 2^{2}.3^{3}.5^{2}.7, but there are more examples outside this table.) Thus the likely rules for σ can be read off from the numerical data with considerable confidence. It is known that σ is a product of local factors. Write

a = ∏ p^{n_{p}}, c_{p} = a/p^{n_{p}}, σ = ∏ σ_{p}, χ_{p} = (\begin{matrix}c_{p} \\ p \end{matrix})_{4} .

Then the numerical data fits the following formula:

σ_{2} = −1 if \begin{cases}n_{2} = 0 and c_{2} ≡ 3, 9, 11 or 13 mod 16, \\ n_{2} = 1, \\ n_{2} = 2 and c_{2} ≡ 1, 9, 11 or 15 mod 16, \end{cases}

and σ_{2} = 1 otherwise. For p odd,

σ_{p} = −1 if n_{p} = 2 and p ≡ 3 mod 4

and σ_{p} = 1 otherwise. This is probably not hard to prove, though tedious.

The following small table, which covers those a divisible only by 2,3 and 5, gives the values of

2^{3−2t}f_{odd}^{2}ω^{−3}a^{1/4}L_{1}(3)

where t is the number of distinct odd primes which divide a. But the integers involved have no particular allure.

\begin{matrix}1 & 0 & 0 & 4·000 & 4·000 & 4·000 & 6·000 \\ 3 & 1 & 0 & 10·000 & 14·000 & 16·000 & 15·000 \\ 9 & 2 & 0 & 12·000 & 15·000 & 24·000 & 24·000 \\ 27 & 3 & 0 & 18·000 & 18·000 & 24·000 & 39·000 \\ 5 & 0 & 1 & 32·000 & 38·000 & 48·000 & 51·000 \\ 25 & 0 & 2 & 48·000 & 58·000 & 68·000 & 75·000 \\ 125 & 0 & 3 & 72·000 & 86·000 & 96·000 & 123·000 \end{matrix}

\begin{matrix}15 & 1 & 1 & 96·000 & 115·000 & 144·000 & 163·500 \\ 75 & 1 & 2 & 145·000 & 184·000 & 204·000 & 240·000 \\ 375 & 1 & 3 & 228·000 & 265·000 & 284·000 & 379·500 \\ 45 & 2 & 1 & 120·000 & 156·000 & 186·000 & 222·000 \\ 225 & 2 & 2 & 192·000 & 247·500 & 264·000 & 312·000 \\ 1125 & 2 & 3 & 300·000 & 327·000 & 402·000 & 486·000 \\ 135 & 3 & 1 & 156·000 & 207·000 & 252·000 & 283·500 \\ 675 & 3 & 2 & 261·000 & 306·000 & 360·000 & 411·000 \\ 3375 & 3 & 3 & 408·000 & 441·000 & 528·000 & 619·500 \end{matrix}

To start the calculation of L_{2}(3), we use f_{2}^{2}(z) = g_{2}^{2}(z) = −\frac{1}{2}℘′(z) and g_{2}^{1}(z) = h_{2}(3) + O(z^{2}), g_{2}^{0} = O(z) near z = 0. But

\begin{matrix}f_{2}^{1}(z) = g_{2}^{1}(z) + π^{−1}ω^{2}ζ(z)g_{2}^{2}(z), \\ f_{2}^{0}(z) = g_{2}^{0}(z) + 2π^{−1}ω^{2}ζ(z)g_{2}^{1}(z) + (π^{−1}ω^{2}ζ(z))^{2}g_{2}^{2}(z). \end{matrix}

Since ℘(z) = z^{−2} + \frac{1}{5}z^{2} + O(z^{6}) near z = 0, the first equation gives

f_{2}^{1}(z) = h_{2}(3) + π^{−1}ω^{2}(z^{−4} − \frac{4}{15}) + O(z^{4})

and therefore

f_{2}^{1}(z) = π^{−1}ω^{2}((℘(z))^{2} − \frac{2}{3}) + h_{2}(3).

The equation for f_{2}^{0}(z) now becomes

f_{2}^{0}(z) = 2π^{−1}ω^{2}z^{−1}h_{2}(3) + π^{−2}ω^{4}(z^{−5} − \frac{1}{3}z^{−1}) + O(z);

and this implies that

f_{2}^{0}(z) = −\frac{1}{2}π^{−2}ω^{4}℘(z)℘′(z) and h_{2}(3) = \frac{1}{6}π^{−1}ω^{2}.

This time we tabulate the values of

2^{3−2t}f_{odd}^{2}π^{2}ω^{−5}a^{1/4}L_{2}(3).

\begin{matrix}1 & 0 & 0 & 0·000 & 0·000 & 4·000 & 8·000 \\ 3 & 1 & 0 & 2·000 & 0·000 & 16·000 & 100·000 \\ 9 & 2 & 0 & 0·000 & 36·000 & 0·000 & 0·000 \end{matrix}

\begin{matrix}27 & 3 & 0 & 18·000 & 0·000 & 0·000 & 36·000 \\ 5 & 0 & 1 & 16·000 & 0·000 & 0·000 & 4·000 \\ 25 & 0 & 2 & 32·000 & 0·000 & 4·000 & 324·000 \\ 125 & 0 & 3 & 0·000 & 0·000 & 0·000 & 196·000 \\ 15 & 1 & 1 & 0·000 & 0·000 & 32·000 & 338·000 \\ 75 & 1 & 2 & 121·000 & 0·000 & 0·000 & 64·000 \\ 375 & 1 & 3 & 0·000 & 0·000 & 0·000 & 722·000 \\ 45 & 2 & 1 & 72·000 & 576·000 & 162·000 & 0·000 \\ 225 & 2 & 2 & 288·000 & 18·000 & 0·000 & 0·000 \\ 1125 & 2 & 3 & 0·000 & 900·000 & 18·000 & 0·000 \\ 135 & 3 & 1 & 0·000 & 0·000 & 0·000 & 18·000 \\ 675 & 3 & 2 & 225·000 & 0·000 & 72·000 & 324·000 \\ 3375 & 3 & 3 & 0·000 & 0·000 & 288·000 & 5202·000 \end{matrix}

The sign of the functional equation is again the product of local factors, but this time the rules are

σ_{2} = −1 if \begin{cases}n_{2} = 0 and c_{2} ≡ 1, 7, 13 or 15 mod 16, \\ n_{2} = 1, \\ n_{2} = 2 and c_{2} ≡ 5, 7, 11 or 13 mod 16, \end{cases}

and σ_{2} = 1 otherwise. For p odd,

σ_{p} = −1 if n_{p} = 2 and p ≡ 3 mod 4

and σ_{p} = 1 otherwise.

6. The cases n = 5 and n = 10. [Not yet written]

REFERENCES

[1] Berndt,B.C., Evans,R.J. and Williams,K.S., Gauss and Jacobi Sums (Wiley, 1998).

[2] Birch,B.J. and Swinnerton-Dyer,H.P.F., Notes on elliptic curves II, J. für reine angew. Math. 218(1965), 79-108.

[3] Pinch,R.G.E. and Swinnerton-Dyer,H.P.F., Arithmetic of diagonal quartic surfaces I, in L-functions and Arithmetic (ed. J.Coates and M.J.Taylor), London Math. Soc. Lecture Note series 153 (Cambridge, 1991).

[4] Shioda,T., The Hodge Conjecture for Fermat Varieties, Math. Ann. 245(1979), 175-184.

[5] Swinnerton-Dyer,H.P.F., The Conjectures of Birch and Swinnerton-Dyer, and of Tate, in Proceedings of a Conference on Local Fields (ed. T.A.Springer) (Springer, 1967).

[6] Washington,L.C., Introduction to Cyclotomic Fields (2nd ed., Springer, 1991).

[7] Weil,A., Jacobi sums as ”Grossencharaktere”, Trans. Amer. Math. Soc. 73(1952),487-495 = Oeuvres Scientifiques II,63-71.

[8] Weil,A., Sommes de Jacobi et charactères de Hecke, Gött. Nachr.(1974), Nr 1 = Oeuvres Scientifiques III,329-342.