MEAN VALUES OF DERIVATIVES OF MODULAR L-SERIES

M. Ram Murty and V. Kumar Murty

§1. Introduction.

Recently, Kolyvagin [4] proved the finiteness of the Tate-Shafarevic group of certain modular elliptic curves over Q. More precisely, let E/Q be a modular elliptic curve and L(s) its associated L-series:

L(s) = ∑_{n=1}^{∞} \frac{a(n)}{n^{s}}.

Set

L_{D}(s) = ∑_{n=1}^{∞} \frac{a(n)}{n^{s}}(\frac{D}{n} )

where (\frac{D}{n} ) is the Legendre symbol. Suppose that L(1) ≠ 0 and there exists a D < 0, such that

(i) L_{D}(s) has a simple zero at s = 1 and (ii) all primes which divide the conductor of E split in the imaginary quadratic field

Q(\sqrt{D}).

Under these conditions, Kolyvagin [4] showed that both E(Q) and III_{E/Q} are finite.

More recently, he extended this theorem to show that if L(s) has a simple zero at s = 1 and there is a D < 0 satisfying (ii) above and L_{D}(1) ≠ 0, then rank E(Q) = 1 and III_{E/Q} is finite.

Previously, Rubin [10] established the finiteness of III_{E/Q} for CM elliptic curves of zero Mordell-Weil rank. The work of Rubin and Kolyvagin represent significant steps toward the resolution of the important conjecture that III_{E/Q} is finite.

The purpose of this paper is to establish the existence of a D < 0 such that L_{D}(s) has a simple zero at s = 1 and all primes dividing the conductor of E split completely in the quadratic field Q(\sqrt{D}). Thus, the result of Kolyvagin can be stated without any hypothesis on quadratic twists of L(s). We prove our theorem by showing that the mean value of a certain expression involving L′_{D}(1) and L′′_{D}(1) is non-zero. More precisely, we show that:

Theorem. Write each D as D_{0}δ^{2}, where D_{0} is a fundamental discriminant and set

R_{D}(X) = L′_{D_{0}}(1) ∑_{d|δ^{2}} \frac{μ̃(d)}{d}(\frac{D_{0}}{d} ) log \frac{Xd}{δ^{4}} + \frac{1}{2}L′′_{D_{0}}(1).

Let

C = \frac{1}{ϕ(4N)} ∑_{n=1}^{∞} \frac{ϕ(4Nn)}{4Nn} \frac{a(n^{2})}{n^{2}},

where ϕ denotes Euler’s function. Then, there is a ν > 0, such that for

Y ≤ X ≤ Y (log Y)^{1+ν},

we have

∑_{{0<−D≤Y \atop D≡1( mod 4N)}} R_{D}(X) = 2CY (log Y)(log \frac{X}{Y}) + O(Y log Y),

as Y →∞.

It is not difficult to see that the constant in the leading term of the asymptotic formula is non-zero. Indeed, if we set

L(s, sym^{2}) = ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{s}},

then

∑_{n=1}^{∞} \frac{ϕ(4Nn)a(n^{2})}{4Nn^{s+1}} = \frac{ϕ(4N)}{4N}L(s + 1, sym^{2})P (s + 1)

where P (s) has an Euler product which converges absolutely for ℜs > 1. Indeed,

P (s) = ∏_{p|̸4N} (1 − \frac{1}{p^{s+1}} \{a(p^{2})(1 − \frac{1}{p^{s−1}} ) + \frac{1}{p^{2s−3}} \}) .

This is seen in the following way:

∑_{n=1}^{∞} \frac{ϕ(4Nn)}{4Nn} \frac{a(n^{2})}{n^{s}}
= ∑_{{b=1 \atop p|b⇒p|4N}}^{∞} ∑_{{m=1 \atop (m,4N)=1}}^{∞} \frac{a(b^{2}m^{2})}{(bm)^{s}} \frac{ϕ(4Nbm)}{4Nbm}
= ∑_{{b=1 \atop p|b⇒p|4N}}^{∞} \frac{a(b^{2})}{b^{s}} \frac{ϕ(4N)}{4N} ∑_{{m=1 \atop (m,4N)=1}}^{∞} \frac{a(m^{2})}{m^{s}} \frac{ϕ(m)}{m}
=\frac{ϕ(4N)}{4N} ∏_{p|4N}(1 − α(p)^{2}p^{−s})^{−1}(1 − p^{1−s})^{−1}(1 − β(p)^{2}p^{−s})^{−1} ∑_{{m=1 \atop (m,4N)=1}}^{∞} \frac{a(m^{2})}{m^{s}} \frac{ϕ(m)}{m}.

Now this last series has an Euler product of which the p-factor is

1 + (1 − \frac{1}{p}) (\frac{a(p^{2})}{p^{s}} + \frac{a(p^{4})}{p^{2}s} + ...) .

This p-factor simplifies to

1 − \frac{1}{p^{s+1}} (a(p^{2}) (1 − \frac{1}{p^{s−1}} ) + \frac{1}{p^{2s−3}} ) .

In particular, P (2) ≠ 0. Since L(2, sym^{2}) ≠ 0 (see [5] or [6]) we deduce that the constant is non-zero. This leads to:

Corollary. There are infinitely many D < 0 such that L_{D}(s) has a simple zero at s = 1 and all primes dividing the conductor of E split completely in the imaginary quadratic field Q(\sqrt{D}).

Proof. The square values of −D contribute O(Y^{1/2} log Y) to the sum. Removing these values of D from the sum and noting that D ≡ 1( mod 4N) is the Heegner condition gives the corollary from Theorem 2.

The theorem and corollary were also established by Bump, Friedberg and Hoffstein [1] utilising the automorphic theory of GSP (4).

Our methods apply also to L series obtained from modular forms of integral weight though we only develop the theorems here for those of weight 2. To fix ideas, let F (z) be a cusp form of weight 2 on Γ_{0}(N) and write

F (z) = ∑_{n=1}^{∞} a(n)e^{2πinz}

be its Fourier expansion at the cusp i∞. Let

L(s) = ∑_{n=1}^{∞} \frac{a(n)}{n^{s}}

be the associated L-series. L(s) satisfies a functional equation:

A^{s}Γ(s)L(s) = wA^{2−s}Γ(2 − s)L(2 − s),

where w = ±1 and A = \sqrt{N}/2π. Let χ_{D}(n) = (\frac{D}{n} ) be a real character mod D. Then, we can consider

L_{D}(s) = ∑_{n=1}^{∞} \frac{a(n)}{n^{s}} (\frac{D}{n} ) .

If D is a fundamental discriminant, then L_{D}(s) satisfies a functional equation:

(A|D|)^{s}Γ(s)L_{D}(s) = wχ_{D}(−N)(A|D|)^{2−s}Γ(2 − s)L_{D}(2 − s).

Hence, if D_{0} is a fundamental discriminant,

L_{D_{0}}(1 + s) = wχ_{D_{0}}(−N)|D_{0}|^{−2s}A^{−2s} \frac{Γ(1 − s)}{Γ(1 + s)}L_{D_{0}}(1 − s).

For convenience, we also define

\frac{1}{L(s)} = ∑_{n=1}^{∞} \frac{μ̃(n)}{n^{s}}.

In case F is a Hecke eigenform, we can write

\frac{1}{L(s)} = ∏_{p}(1 − \frac{a(p)}{p^{s}} + \frac{1}{p^{2s−1}} ).

Any D can be written as, D = D_{0}r^{2}, where D_{0} is a fundamental discriminant and

L_{D}(s) = L_{D_{0}}(s) ∑_{d|r^{2}} \frac{μ̃(d)}{d^{s}} (\frac{D_{0}}{d} ) .

We will also set

f_{Y} (n, s; a) = ∑_{{0<−D_{0}≤Y \atop D_{0}≡a( mod 4N)}} (\frac{D_{0}}{n} ) |D_{0}|^{s}, f_{Y} (n; a) = f_{Y} (n, 0; a)

where the summation is over fundamental discriminants.

We begin by giving an outline of the proof. §2. Outline of proof.

Consider

∑_{n≤X} \frac{a(n)}{n}(\frac{D}{n} ) log \frac{X}{n} = \frac{1}{2πi} ∫_{(σ)} L_{D}(1 + s)\frac{X^{s}}{s^{2}} ds

for σ > 1/2. Moving the line of integration to ℜs = −η, we obtain

L′_{D}(1) = ∑_{n≤X} \frac{a(n)}{n}(\frac{D}{n} ) log \frac{X}{n} − \frac{1}{2πi} ∫_{(−η)} L_{D}(1 + s)\frac{X^{s}}{s^{2}} ds.

Each D can be written uniquely as D_{0}δ^{2} where D_{0} is a fundamental discriminant. Hence, writing

L_{D}(s) = L_{D_{0}}(s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{s}}(\frac{D_{0}}{d} )

and applying the functional equation to L_{D_{0}}(s) we obtain

L′_{D}(1) = ∑_{n≤X} \frac{a(n)}{n}(\frac{D}{n} ) log \frac{X}{n}

−\frac{wχ_{D_{0}}(−N)}{2πi} ∫_{(−η)} |D_{0}|^{−2s}L_{D_{0}}(1 − s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{1+s}} (\frac{D_{0}}{d} )\frac{Γ(1 − s)}{Γ(1 + s)} \frac{X^{s}A^{−2s}}{s^{2}} ds.

Suppose for now that N = 1. Since D_{0} < 0, χ_{D_{0}}(−1) = −1. If η > 1/2, we can write L_{D_{0}}(1 − s) as a Dirichlet series, so that if we sum the above expression over 0 < −D ≤ Y, D ≡ 1( mod 4), that is over δ^{2} ≤ Y, δ odd and fundamental discriminants 0 < −D_{0} ≤ Y/δ^{2}, D_{0} ≡ 1( mod 4), we obtain

∑_{{0<−D≤Y \atop D≡1( mod 4)}} L′_{D}(1) = ∑_{n≤X} \frac{a(n)}{n}f̃_{Y} (n) log \frac{X}{n}

+\frac{w}{2πi} ∑_{{δ^{2}≤Y \atop δ odd}} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n)}{n^{1−s}} f_{Y/δ^{2}}(nd, −2s; δ̄^{2})\frac{Γ(1 − s)}{Γ(1 + s)} \frac{X^{s}A^{−2s}}{s^{2}} ds,

where

f̃_{Y} (n) = ∑_{{0<−D≤Y \atop D≡1( mod 4)}} (\frac{D}{n} ),

and c̄ denotes the inverse residue class of c( mod 4N). Let us first consider the sum. We split it into two parts according as n is a square or not a square. When n is a square, it is easily seen that the sum is

\frac{Y}{2} ∑_{n^{2}≤X} \frac{a(n^{2})}{n^{2}} \frac{ϕ(4n)}{4n} log \frac{X}{n^{2}} ∼ \frac{Y log X}{2} ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{2}} \frac{ϕ(4n)}{4n}

as X→∞. When n is not a square, we have by the Polya-Vinogradov inequality, an error term of

∑_{n≤X} \frac{|a(n)|}{\sqrt{n}}(log n) log \frac{X}{n} .

Rankin [9] has shown that

∑_{n≤X} \frac{|a(n)|}{\sqrt{n}} ≪ X(log^{−ϵ} X)

for some ϵ > 0. Therefore, the total error from the sum does not exceed

O(X log^{1−ϵ} X).

The analysis of the integral is slightly more complicated owing to the presence of the fundamental discriminants. The Polya-Vinogradov inequality is no longer applicable. However, the inequality of Jutila and Fainleib-Saparnijazov (Lemma 1) and the lemma of §4 can be applied. One splits the integral into the contribution arising when nd is a square and when nd is not a square. The ideas for the detailed analysis appear in sections 4 and 5. The error term comes out to be

d^{1/2}X^{−1/2}Y^{3/2}δ^{−3} log X log log X.

Summing this, gives a total error of

∑_{δ^{2}≤Y} ∑_{d|δ^{2}} \frac{|μ̃(d)|}{d}d^{1/2}δ^{−3}Y^{3/2}X^{−1/2} log X log log X ≪ X^{−1/2}Y^{3/2} log X log log X.

The contribution when nd is a square comes out to be

−\frac{Y}{2} log \frac{X}{Y^{2}} ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{2}} \frac{ϕ(4n)}{4n} + O(Y).

Therefore, the final result is

∑_{{0<−D≤Y \atop D≡1( mod 4)}} L′_{D}(1) = Y log Y ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{2}} \frac{ϕ(n)}{n} + O(X log^{1−ϵ} X)

+O(X^{−1/2}Y^{3/2} log X log log X).

Choosing X = Y log^{ν} Y for some ν < ϵ, we obtain the desired asymptotic formula. When N > 1, the analysis is similar, except that we replace Jutila’s inequality by one of Fainleib and Saparnijazov (Lemma 1).

The above was our original approach. However, it is somewhat technically simpler to consider

\frac{1}{2πi} ∫_{(σ)} L_{D_{0}}(1 + s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{1−s}} (\frac{D_{0}}{d} )\frac{X^{s}}{s^{3}} δ^{−4s}ds.

The advantage here is that when we apply the functional equation, we obtain L_{D}(1 − s) in the integrand. Hence, we can apply the Polya-Vinogradov inequality and Rankin’s result to estimate the integral. This transfers the tedious estimations to the sum. Moreover, the s^{3} factor removes any subtle convergence questions and provides for a swifter arrival to the corollary. We have therefore taken this approach here. We relegate the analysis of the mean values of the derivatives themselves to a future paper.

The next two sections establish the requisite lemmas to estimate the sums we will encounter.

§3. Lemmas.

Lemma 1. (Fainleib and Saparnijazov, Jutila)

∑_{n≤X} ′ | ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{n}) |^{2} ≪ MXY log^{2} X

where the implied constant is absolute and the sum over n ≤ X is over non-squares.

Lemma 2. Let d and a be fixed integers, with a ≡ 1( mod 4). Then,

| ∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n}f_{Y} (nd; a)| ≪ (Ud)^{1/2}Y^{1/2} log Y log(Ud).

Proof. By partial summation and Lemma 1,

∑_{m≤U} ′ \frac{1}{\sqrt{m}}| ∑_{{h≤Y \atop h≡a( mod 4N)}} (\frac{h}{m})|^{2} ≪ U^{1/2}Y log^{2} U.

Let χ_{0}^{(a)} denote the principal character mod a. The sum in question is

∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n} ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{nd}) ∑_{j^{2}|D} μ(j)

= ∑_{j^{2}≤Y} μ(j) ∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n}χ_{0}^{(j)}(nd) ∑_{{h≤Y /j^{2} \atop h≡aj̄^{2}( mod 4N)}} (\frac{h}{nd})

≪ ∑_{j^{2}≤Y} ( ∑_{n≤U} \frac{|a(n)|^{2}}{n^{3/2}} )^{1/2} ( ∑_{n≤U,nd≠b^{2}} \frac{1}{\sqrt{n}}| ∑_{{h≤Y /j^{2} \atop h≡aj̄^{2}( mod 4N)}} (\frac{h}{nd})|^{2})^{1/2}

≪ U^{1/4}d^{1/4} ∑_{j^{2}≤Y} ( ∑_{m≤Ud}′ \frac{1}{\sqrt{m}}| ∑_{{h≤Y /j^{2} \atop h≡aj̄^{2}( mod 4N)}} (\frac{h}{m})|^{2})^{1/2}

≪ U^{1/2}d^{1/2}Y^{1/2}(log Y)(log Ud),

as desired. Lemma 3. For ℜs = −η, 0 < η < 1/2, a ≡ 1( mod 4), we have

∑_{n≤U,nd≠b^{2}} \frac{a(n)χ_{0}^{(j)}(nd)}{n^{1−s}} ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{nd}) ≪ U^{1/2−η}d^{1/2}Y^{1/2} log Ud.

Proof. We apply Cauchy’s inequality:

\begin{matrix}( ∑_{n≤U} \frac{|a(n)|^{2}}{n^{3/2+η}} ) & 1/2 & ( ∑_{n≤U,nd≠b^{2}} \frac{1}{n^{1/2+η}} | ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{nd})|^{2}) & 1/2 \\ ≪ d^{1/2}U^{1/2−η}Y^{1/2} log Ud \end{matrix}

by partial summation. Lemma 4. Under the same conditions as lemma 3,

∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} χ_{0}^{(j)}(nd) ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{nd})h^{−2s} ≪ (|s| + 1)d^{1/2}U^{1/2−η}Y^{1/2+2η} log Ud.

Proof. Apply partial summation to lemma 3. Lemma 5. If ℜs = −η, 0 < η < 1/2, a ≡ 1( mod 4) then

∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} f_{Y} (nd, −2s; a) ≪ (|s| + 1)d^{1/2}U^{1/2−η}Y^{1/2+2η} log Y log Ud.

Proof. The sum in question is

∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{nd}) ∑_{j^{2}|D} μ(j)D^{−2s}

= ∑_{j^{2}≤Y} μ(j)j^{−4s} ∑_{n≤U,nd≠b^{2}} \frac{a(n)χ_{0}^{(j)}(nd)}{n^{1−s}} ∑_{{h≤Y /j^{2} \atop h≡aj̄^{2}( mod 4N)}} (\frac{h}{nd})h^{−2s}

≪ ∑_{j^{2}≤Y} j^{4η}(|s| + 1)(d^{1/2}U^{1/2−η} log Ud)\frac{Y^{1/2+2η}}{j^{1+4η}}

by lemma 4. Summing over j gives the desired result.

Putting the above lemmas together, proves: Lemma 6. For ℜs = −δ, 0 < δ < 1/2, a ≡ 1( mod 4) k ≥ 3,

∫_{(−δ)} ∑_{n≤U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} f_{Y} (nd, −2s; a)\frac{x^{s}}{s^{k}} ds ≪ x^{−δ}d^{1/2}U^{1/2−δ}Y^{1/2+2δ} log Y log Ud.

We now proceed to handle the terms corresponding to n > U. Lemma 7. For ℜs = −η, η > 1/2,

∑_{n>U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} χ_{0}^{(j)}(nd) ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{nd}) ≪ U^{1/2−η}d^{1/2}Y^{1/2} log Ud.

Proof. By Cauchy’s inequality, the sum is bounded by

( ∑_{n>U} \frac{|a(n)|^{2}}{n^{3/2+η}} )^{1/2} ( ∑_{n>U,nd≠b^{2}} \frac{1}{n^{1/2+η}} | ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{nd})|^{2})^{1/2}

≪ U^{1/4−η/2}d^{1/4+η/2} ( ∑_{m>Ud}′ \frac{1}{m^{1/2+η}} | ∑_{{h≤Y \atop h≡a( mod M)}} (\frac{h}{m})|^{2})^{1/2}

≪ U^{1/4−η/2}d^{1/4+η/2} (∫_{Ud}^{∞} \frac{Y log^{2} t}{t^{1/2+η}}dt)^{1/2}

≪ U^{1/2−η}d^{1/2}Y^{1/2} log Ud,

as desired. Lemma 8. Under the same conditions as lemma 7,

∑_{n>U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} χ_{0}^{(j)}(nd) ∑_{h≤Y} (\frac{h}{nd})h^{−2s} ≪ (|s| + 1)Y^{1/2+2η}d^{1/2}U^{1/2−η} log Ud.

Proof. Apply partial summation to lemma 7. Lemma 9. If ℜs = −η, η > 1/2, a ≡ 1( mod 4), then

∑_{n>U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} f_{Y} (nd, −2s; a) ≪ (|s| + 1)d^{1/2}U^{1/2−η}Y^{1/2+2η} log Y log Ud.

Proof. The proof is analogous to lemma 5, except we use lemma 8 instead of lemma 4.

Therefore, we deduce by putting these lemmas together, Lemma 10. For ℜs = −η, η > 1/2, a ≡ 1( mod 4), k ≥ 3

∫_{(−η)} ∑_{n>U,nd≠b^{2}} \frac{a(n)}{n^{1−s}} f_{Y} (nd, −2s; a)\frac{x^{s}}{s^{k}} ds ≪ x^{−η}d^{1/2}U^{1/2−η}Y^{1/2+2η} log Y log Ud.

These are the series of lemmas that we need. The next section establishes a lemma which refines the above estimates.

§4. Further lemmas.

We begin by proving

Lemma 11. Let a ≡ 1( mod 4) and Y ≤ X ≤ Y (log Y)^{1+ν}. Set

r_{D}(X) = L′_{D}(1) log X + \frac{1}{2}L′′_{D}(1).

Then,

∑_{{0<−D≤Y \atop D≡a( mod 4N)}} r_{D}(X) ≪ Y log^{2} Y.

Proof. Consider,

\frac{1}{2πi} ∫_{(σ)} L_{D}(1 + s)\frac{X^{s}}{s^{3}} ds, σ > 1/2.

This is

\frac{1}{2} ∑_{n≤X} \frac{a(n)}{n}(\frac{D}{n} ) log^{2} \frac{X}{n} .

By the Phragmén-Lindelöf theorem [8],

|L_{D}(1 + s)| ≪ (|t| + 2)^{1/2−σ}(log(|t| + 2))^{2}

uniformly for −1/2 − ϵ ≤ σ ≤ 1/2 + ϵ. We can therefore move the line of integration to ℜs = −η ≥ −1/2 − ϵ.

Moving the line of integration to ℜs = −η, we obtain

\frac{1}{2} ∑_{n≤X} \frac{a(n)}{n}(\frac{D}{n} ) log^{2} \frac{X}{n} = r_{D}(X) + \frac{1}{2πi} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n)}{n^{1+s}} (\frac{D}{n} )\frac{X^{s}}{s^{3}} ds.

Writing

L_{D}(1 + s) = L_{D_{0}}(1 + s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{1+s}} (\frac{D_{0}}{d} )

where D_{0} is a fundamental discriminant, we can rewrite the integral as

∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n)}{n^{1+s}} (\frac{D_{0}}{nd} )\frac{(X/d)^{s}}{s^{3}} ds.

Applying the functional equation to L_{D_{0}}(1 + s), we get the above as

\frac{wχ_{D_{0}}(−N)}{2πi} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n)}{n^{1−s}} (\frac{D_{0}}{nd} )|D_{0}|^{−2s}A^{−2s} \frac{Γ(1 − s)}{Γ(1 + s)} \frac{(X/d)^{s}}{s^{3}} ds.

We sum this over δ^{2} ≤ Y, 0 < −D_{0} ≤ Y/δ^{2} D_{0}δ^{2} ≡ a( mod 4N). Let

f̃_{Y} (n, s; a) = ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{n} )|D|^{s}, D unrestricted
f_{Y} (n, s; a) = ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{n} )|D|^{s}, D fundamental
f̃_{Y} (n; a) = f̃_{Y} (n, 0; a),  f_{Y} (n; a) = f_{Y} (n; 0, a).

Then,

(⋆)  \frac{1}{2} ∑_{n≤X} \frac{a(n)}{n}f̃_{Y} (n; a) log^{2} \frac{X}{n} = ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} r_{D}(X)+

\frac{1}{2πi} ∑_{{δ^{2}≤Y \atop (δ,4N)=1}} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n)}{n^{1−s}} f_{Y/δ^{2}}(nd, −2s; aδ̄^{2})\frac{Γ(1 − s)}{Γ(1 + s)}A^{−2s} \frac{(X/d)^{s}}{s^{3}} ds.

First we deal with the sum on the left side of (⋆). As noted in section 2, the contribution from non-squares is

≪ X(log X)^{1−ϵ} ≪ Y (log Y)^{2},

if X < Y (log Y)^{1+ν}, for ν < ϵ. The contribution from squares is

≪ Y log^{2} X.

Now we deal with the integral on the right side of (⋆). The integral is first broken up according as nd is a square or not a square. If nd is not a square, then splitting the series at n ≤ Y and moving the integral to ℜs = −η_{1}, 0 < η_{1} < 1/2, we can utilise Lemma 6 to get an estimate of

∑_{δ^{2}≤Y} ∑_{d|δ^{2}} \frac{|μ̃(d)|}{d}d^{1/2} \frac{Y}{δ^{1+4η_{1}}} (\frac{Y}{X})^{η_{1}} log^{2} Y ≪ Y (Y/X)^{η_{1}} log^{2} Y.

The series corresponding to n ≥ Y is similarly estimated by moving to ℜs = −η_{2}, η_{2} > 1/2 and using lemma 10. The final contribution when nd is not a square is therefore

Y \{(\frac{Y}{X} )^{η_{1}} + (\frac{Y}{X} )^{η_{2}}\} log^{2} Y

This term is ≪ Y log^{2} Y if Y ≤ X. Now we consider the contribution when nd is a square.

Let us therefore set

F_{d}(s) = ∑_{n=1}^{∞} \frac{a(dn^{2})}{(dn^{2})^{s}},

when d is squarefree. We note that dn is a square if and only if d_{0}n is a square, where d = d_{0}d_{1}^{2}, d_{0} squarefree. It therefore suffices to consider F_{d}(s) when d is squarefree.

Let us first consider the case d = p is prime. Then

F_{p}(s) = ∑_{{n=1, \atop (n,p)=1,α≥0}}^{∞} \frac{a(p^{2α+1})}{p^{(2α+1)s}} \frac{a(n^{2})}{n^{2s}},

which is

F_{p}(s) = L(2s, sym^{2}) (∑_{α=0}^{∞} \frac{a(p^{2α})}{p^{2αs}})^{−1} (∑_{α=0}^{∞} \frac{a(p^{2α+1})}{p^{(2α+1)s}}) .

Using, |a(n)| ≤ n^{1/2}d(n), we estimate

(∑_{α=0}^{∞} \frac{a(p^{2α})}{p^{2αs}})^{−1} = (1 − \frac{α(p)^{2}}{p^{2s}} ) (1 − \frac{α(p)β(p)}{p^{2s}}) (1 − \frac{β(p)^{2}}{p^{2s}} )

≪ (1 + \frac{1}{p^{2σ−1}} )^{3}

≪ 8

for σ ≥ 1/2. Now,

∑_{α=0}^{∞} \frac{a(p^{2α+1})}{p^{(2α+1)s}} ≪ ∑_{α=0}^{∞} \frac{2α + 1}{p^{(2α+1)(σ−1/2)}} ≪ 2p^{σ−1/2} ∑_{α=0}^{∞} \frac{α + 1}{p^{(2σ−1)(α+1)}} .

Using,

∑_{n=0}^{∞}(n + 1)t^{n+1} = t/(1 − t)^{2},

we obtain an estimate of

16p^{1/2−σ}(1 − p^{1−2σ})^{−2}, σ > 1/2.

Thus, for σ ≥ 3/4,

|F_{p}(s)| ≪ 128p^{1/2−σ}|L(2s, sym^{2})|.

It is now clear that if d = p_{1}p_{2}...p_{k} , then

F_{d}(s) = ∑_{{n=1 \atop (n,d)=1;α_{1},...,α_{k}≥0}}^{∞} \frac{a(p_{1}^{2α_{1}+1})...a(p_{k}^{2α_{k}+1})}{p_{1}^{(2α_{1}+1)s}...p_{k}^{(2α_{k}+1)s}} \frac{a(n^{2})}{n^{2s}}

and by the method indicated above,

|F_{d}(s)| ≪ (128)^{ν(d)}d^{1/2−σ}|L(2s, sym^{2})|,

for σ ≥ 3/4, where ν(d) is the number of prime factors of d. It follows that for any d (squarefree or not), we have

|F_{d}(s)| ≪ (128)^{ν(d)}|L(2s, sym^{2})|, σ ≥ 3/4.

We are now ready to estimate the integral when nd is a square. By Jutila’s result [3, Lemma 1] on f(n, s) when n is a square, we find the integral is

∫_{(−η)} ∑_{{n=1 \atop nd=b^{2}}}^{∞} \frac{a(n)}{n^{1−s}} \{\frac{(Y/δ^{2})^{1−2s}}{1 − 2s} + O ((|s| + 1)\frac{Y^{1/2+2η}}{δ^{1+4η}} )\} A^{−2s} \frac{Γ(1 − s)}{Γ(1 + s)} \frac{(X/d)^{s}}{s^{3}} ds.

We move this integral to η = 1/ log Y and evaluate it using the above estimate for F_{d}(s). The error term presents no problem. The main term is

≪ \frac{Y}{δ^{2}} (d/X)^{η}(128)^{ν(d)}(log Y)^{2}.

Summing over d|δ^{2} and δ^{2} ≤ Y, we obtain a total estimate of

≪ Y ∑_{δ^{2}≤Y} \frac{1}{δ^{2}} ∑_{d|δ^{2}} \frac{|μ̃(d)|}{d}(d/X)^{η}(128)^{ν(d)} ≪ Y (log Y)^{2}.

This completes the proof of lemma 11.

Utilising Lemma 11, we can now derive an alternate estimate for the sum

∑_{n≤X} \frac{a(n)}{n}f̃_{Y} (n; a) log^{2} \frac{X}{n} .

Lemma 12. If Y ≤ X < Y (log Y)^{1+δ}, a ≡ 1( mod 4), then

∑_{n≤X} \frac{a(n)}{n} ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{n} ) log^{2} \frac{X}{n} ≪ Y log^{2} Y,

and

∑_{{n≤X \atop (n,a)=1}} \frac{a(n)}{n} ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{n} ) log^{2} \frac{X}{n} ≪ Y log^{2} Y.

Remark. Note that in these sums we are not restricting to fundamental discriminants.

Proof. The first inequality follows from the previous analysis. The second assertion is tantamount to considering

∑_{n≤X} \frac{a(n)}{n} ∑_{0<−D≤Y} (\frac{a^{2}D}{n} ) log \frac{X}{n} .

In the previous analysis, this means that we consider those δ^{2} ≤ Y satisfying a|δ. It is seen from the previous discussion that we in fact get estimates of the form

\frac{Y log^{2} Y}{a}

for the desired sums.

Lemma 13 If Y ≤ X < Y (log Y)^{1+δ}, then

∑_{n≤X} \frac{a(n)}{n} ∑_{{0<−D≤Y \atop D≡a( mod 4N)}} (\frac{D}{nd}) log^{2} \frac{X}{n} ≪ d^{1/2}Y log^{2} Y.

Proof. This follows easily from the previous analysis. If before summing over 0 < −D ≤ Y, we multiply by (\frac{D}{d} ) and then sum, all of the previous estimates of section 2 apply and we obtain the above result.

We apply this lemma in the following way.

Lemma 14. For all X satisfying Y (log Y)^{1+ν} ≥ X ≥ Y log^{−B} Y,

∑_{n≤X} ′ \frac{a(n)}{n}f_{Y} (n; c) log^{2} \frac{X}{n} ≪_{B} X^{1/2}Y^{1/2} log X log log Y.

Proof. We have

∑_{n≤X} ′ \frac{a(n)}{n}f_{Y} (n; c) log \frac{X}{n} = ∑_{j^{2}≤Y} μ(j) ∑_{{n≤X \atop (n,j)=1}} \frac{a(n)}{n} log^{2} \frac{X}{n} ∑_{{h≤Y /j^{2} \atop h≡cj̄^{2}( mod 4N)}} (\frac{h}{n}).

For j ≤ log^{B} Y, we estimate by lemma 1 to get

X^{1/2}Y^{1/2} log X log log Y.

For j > log^{B} Y, we use the second part of lemma 12 to get

∑_{j>log^{B} Y} \frac{Y}{j^{2}} log^{2} Y ≪ Y log^{−B+2} Y.

Since X ≥ Y log^{−B} Y, this proves the lemma.

Using the above lemma, we obtain: Lemma 15. For X satisfying Y (log Y)^{1+ν} ≥ X ≥ Y log^{−B} Y,

∑_{{n≤X \atop nd≠b^{2}}} \frac{a(n)}{n}f_{Y} (nd; c) log^{2} \frac{X}{n} ≪ d^{1/2}X^{1/2}Y^{1/2} log X log log X.

Proof. As before, we have that the above sum is

∑_{{j^{2}≤Y \atop (j,4N)=1}} μ(j) ∑_{{n≤X \atop (n,j)=1}} \frac{a(n)}{n} log^{2} \frac{X}{n} ∑_{{h≤Y /j^{2} \atop h≡cj̄^{2}( mod 4N)}} (\frac{h}{nd}).

By lemma 2, we get for j ≤ log^{B} X, an estimate of

d^{1/2}X^{1/2}Y^{1/2} log X log log X.

For j > log^{B} Y, we apply lemma 13 to obtain an estimate of

d^{1/2}Y log^{−B+2} Y.

This proves the lemma. §5. The main theorem.

Now consider

\frac{1}{2πi} ∫_{(σ)} L_{D_{0}}(1 + s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{1−s}} (\frac{D_{0}}{d} )\frac{X^{s}}{s^{3}} δ^{−4s}ds.

This is

\frac{1}{2} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∑_{n≤Xd/δ^{4}} \frac{a(n)}{n}(\frac{D_{0}}{nd} ) log^{2} \frac{Xd}{nδ^{4}} .

On the other hand, moving the line of integration to the left and picking up the residue at s = 0, we obtain

R_{D}(X) + \frac{1}{2πi} ∫_{(−η)} L_{D_{0}}(1 + s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{1−s}} (\frac{D_{0}}{d} )\frac{X^{s}}{s^{3}} δ^{−4s}ds,

where

R_{D}(X) = L′_{D_{0}}(1) ∑_{d|δ^{2}} \frac{μ̃(d)}{d}(\frac{D_{0}}{d} ) log \frac{Xd}{δ^{4}} + \frac{1}{2}L′′_{D_{0}}(1).

Writing the functional equation for L_{D_{0}}(s) we obtain that the above integral is

\frac{wχ_{D_{0}}(−N)}{2πi} ∫_{(−η)} |D|^{−2s}L_{D}(1 − s)\frac{Γ(1 − s)}{Γ(1 + s)}A^{−2s} \frac{X^{s}}{s^{3}} ds.

Let Y be such that Y log^{−B} Y ≤ X ≤ Y (log Y)^{1+ν}. Now we sum the entire expression above over δ^{2} ≤ Y, 0 < −D_{0} ≤ Y/δ^{2}, D_{0}δ^{2} ≡ 1( mod 4N) and obtain the expression:

∑_{{0<−D≤Y \atop D≡1( mod 4N)}} R_{D}(X) = \frac{1}{2} ∑_{{δ^{2}≤Y \atop (δ,4N)=1}} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∑_{n≤Xd/δ^{4}} \frac{a(n)}{n}f_{Y/δ^{2}}(nd; δ̄^{2}) log^{2} \frac{Xd}{nδ^{4}}

+\frac{1}{2πi} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n)}{n^{1−s}} f̃_{Y} (n, −2s; 1)\frac{Γ(1 − s)}{Γ(1 + s)}A^{−2s} \frac{X^{s}}{s^{3}} ds,

for η > 1/2. We split the first sum into the contribution when nd is a perfect square and when nd is not a perfect square. In the latter case, we estimate the sum as follows. For δ > log^{A} Y the estimate is O(Y) by using lemma 2 and partial summation. For δ < log^{A} Y, we have by lemma 15, an estimate of

≪ X^{1/2}Y^{1/2} log X log log X.

The integral is easily estimated by Polya-Vinogradov and Rankin. Truncating the series at U, we move the integral involving the part n ≤ U to ℜs = −η_{1} where η_{1} < 1. The part corresponding to n > U is moved to ℜs = −η_{2} where η_{2} > 1. Choosing U = Y^{2}/X, we obtain the error from the integral as

≪ X^{−1}Y^{2}(log Y)^{1−ν}.

Choosing X = Y/ log^{ϵ} Y for some sufficiently small ϵ yields a total error of o(Y log Y).

The contribution from the squares is easily handled. It is

\frac{1}{2πi} ∫_{(−η)} ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{2(1−s)}} ∑_{D} |D|^{−2s} \frac{Γ(1 − s)}{Γ(1 + s)}X^{s}A^{−2s} \frac{ds}{s^{3}}

where the sum over D is over

0 < −D ≤ Y, D ≡ 1(mod 4N), (D, n) = 1.

The sum over D is asymptotic to

\frac{1}{ϕ(4N)} \frac{ϕ(4Nn)}{4Nn} \frac{Y^{1−2s}}{1 − 2s}.

Inserting this into the integral, we find that it is

\frac{Y}{ϕ(4N)} ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{2}} \frac{ϕ(4Nn)}{4Nn} \frac{1}{2πi} ∫_{(−η)} (\frac{Xn^{2}}{A^{2}Y^{2}} )^{s} \frac{Γ(1 − s)}{Γ(1 + s)} \frac{1}{1 − 2s} \frac{ds}{s^{3}}.

Now using the expansion

\frac{Γ(1 − s)}{Γ(1 + s)} \frac{1}{1 − 2s} = 1 + (2γ + 2)s + (2γ^{2} + 4γ + 4)s + ...

and

(\frac{Xn^{2}}{A^{2}Y^{2}} )^{s} = 1 + (log \frac{Xn^{2}}{A^{2}Y^{2}} )s + \frac{1}{2}(log^{2} \frac{Xn^{2}}{A^{2}Y^{2}} )s^{2} + ...

we find that it is

−\frac{Y}{ϕ(4N)} ∑_{n=1}^{∞} \frac{a(n^{2})}{n^{2}} \frac{ϕ(4Nn)}{4Nn}\{\frac{1}{2} log^{2} \frac{X}{Y^{2}} + (log \frac{X}{Y^{2}} )(log \frac{n^{2}}{A^{2}} )+

\frac{1}{2} log^{2} \frac{n^{2}}{A^{2}} + (2γ + 2) log \frac{Xn^{2}}{A^{2}Y^{2}} + 2γ^{2} + 4γ + 4\} + O(Y).

Next, we consider the sum. It is

∑_{δ} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∑_{n≤Xdδ^{−4};nd=m^{2}} \frac{a(n)}{n}(∑_{D_{0}} 1) log \frac{Xd}{nδ^{4}}

where the sum over δ is over

δ^{2} ≤ Y, (δ, 4N) = 1

and the sum over D_{0} is over

0 < −D_{0} ≤ \frac{Y}{δ^{2}} , D_{0} ≡ δ̄^{2}(mod 4N), (D_{0}, nd) = 1, D_{0} ≡ 1(mod 4), D_{0} fundamental.

The sum over D_{0} can be expressed as

∑_{D_{0}≤Y/δ^{2}} μ^{2}(D_{0})χ_{0}^{(nd)}(D_{0})\frac{1}{ϕ(4N)} ∑_{ψ mod 4N} ψ̄(δ̄^{2})ψ(D_{0})

and this is equal to

∑_{ψ mod 4N} ψ̄(δ̄^{2})\frac{1}{ϕ(4N)} ∑_{D_{0}≤Y/δ^{2}} μ^{2}(D_{0})χ_{0}^{(nd)}(D_{0})ψ(D_{0}).

Now we consider the inner sum. It is

\frac{1}{ϕ(4N)} ∑_{D_{0}≤Y/δ^{2}}( ∑_{a^{2}|D_{0}} μ(a))χ_{0}^{(nd)}(D_{0})ψ(D_{0})

and rearranging, we find it is

\frac{1}{ϕ(4N)} ∑_{a≤\sqrt{Y} /δ} μ(a)χ_{0}^{(nd)}(a^{2})ψ(a^{2}) ∑_{h≤Y/δ^{2}a^{2}} χ_{0}^{(nd)}(h)ψ(h).

If ψ is nontrivial, the innermost sum is over a non-trivial character of conductor dividing 4N. Thus, by Polya-Vinogradov, it is O(1). The whole sum is then

O(\frac{Y ^{\frac{1}{2}}}{δ} \frac{1}{ϕ(4N)}).

On the other hand, if ψ is the trivial character mod 4N, we get

\frac{1}{ϕ(4N)} ∑_{a≤\sqrt{Y} /δ} μ(a)χ_{0}^{(nd)}(a^{2})χ_{0}^{(4N)}(a^{2})\frac{Y}{δ^{2}a^{2}} \frac{ϕ(4Nnd)}{4Nnd}.

Inserting this information into our big sum, we find that it is

\frac{1}{2} ∑_{δ^{2}≤Y} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∑_{n≤Xdδ^{−4};nd=m^{2}} \frac{a(n)}{n} log^{2} \frac{Xd}{nδ^{4}} \frac{1}{ϕ(4N)} \frac{Y}{δ^{2}} \frac{6}{π^{2}} ∏_{p|4Nnd}(1 − \frac{1}{p^{2}} )^{−1} \frac{ϕ(4Nnd)}{4Nnd}

and this simplifies to

\frac{3Y}{π^{2}ϕ(4N)} ∑_{δ^{2}≤Y} \frac{1}{δ^{2}} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∑_{n≤Xdδ^{−4};nd=m^{2}} \frac{a(n)}{n} log^{2} \frac{Xd}{nδ^{4}} ∏_{p|4Nnd}(1 + \frac{1}{p})^{−1}.

Now the sum over n is analysed as follows. (Let us ignore the factor of ∏_{p|4Nnd}(1+1/p)^{−1} for the moment.) Then,

\frac{1}{2} ∑_{{n≤X \atop nd=m^{2}}} \frac{a(n)}{n} log^{2} \frac{X}{n} = \frac{1}{2πi} ∫_{(2)} G_{d}(1 + s)\frac{X^{s}}{s^{3}} ds

where

G_{d}(s) = ∑_{{n=1 \atop nd=m^{2}}} \frac{a(n)}{n^{s}}.

By the same argument that was used for F_{d}(s), one has that

|G_{d}(s)| ≪ c^{ν(d)}|L(2s, sym^{2})|,

for ℜs ≥ 3/4 (say). Thus,

\frac{1}{2} ∑_{{n≤X \atop nd=m^{2}}} \frac{a(n)}{n} log^{2} \frac{X}{n} = \frac{1}{2}G_{d}(1) log^{2} X + G′_{d}(1) log X + \frac{1}{2}G′′_{d}(1) + o(1),

as X→∞. Inserting the main term into our sum, we find it is

\frac{6Y}{π^{2}ϕ(4N)} ∑_{δ^{2}≤Y} ∑_{d|δ^{2}} \frac{μ̃(d)}{d}\{\frac{1}{2} log^{2} \frac{Xd}{δ^{4}} ∑_{nd=m^{2}} \frac{a(n)}{n} ∏_{p|4Nnd}(1 + \frac{1}{p})^{−1}

−(log \frac{Xd}{δ^{4}}) ∑_{nd=m^{2}} \frac{a(n) log n}{n} ∏_{p|4Nnd}(1 + \frac{1}{p})^{−1} + \frac{1}{2} ∑_{nd=m^{2}} \frac{a(n) log^{2} n}{n} ∏_{p|4Nnd}(1 + \frac{1}{p})^{−1}\}.

All but the first term give an amount of O(Y log X). The first term is

\frac{3Y}{π^{2}ϕ(4N)}(log^{2} X) ∑_{δ^{2}≤Y} \frac{1}{δ^{2}} ∑_{d|δ^{2}} \frac{μ̃(d)}{d} ∑_{nd=m^{2}} \frac{a(n)}{n} ∏_{p|4Nm}(1 + \frac{1}{p})^{−1}.

Now using the definition of μ̃, and making use of the fact

∑_{n=1}^{∞} \frac{a(n)}{n^{s}}(\frac{δ^{2}}{n} ) = L(s) ∑_{d|δ^{2}} \frac{μ̃(d)}{d^{s}}

we obtain the identity

a(n)(\frac{δ^{2}}{n} ) = ∑_{{d|δ^{2} \atop dm=n}} μ̃(d)a_{m}.

Therefore, we find this is

\frac{3Y (log^{2} X)}{π^{2}ϕ(4N)} ∑_{δ^{2}≤Y} \frac{1}{δ^{2}} ∑ \frac{a(m^{2})}{m^{2}}(\frac{δ^{2}}{m^{2}} ) ∏_{p|4Nm^{2}}(1 + \frac{1}{p})^{−1}

and this is

\frac{3Y (log^{2} X)}{π^{2}ϕ(4N)} ∑_{m=1}^{∞} \frac{a(m^{2})}{m^{2}} ∏_{p|4Nm^{2}}(1 + \frac{1}{p})^{−1} ∑_{δ^{2}≤Y,(δ,4Nm)=1} \frac{1}{δ^{2}}

which when we extend the summation over δ to infinity, as we may, simplifies to

\frac{1}{2} \frac{Y (log^{2} X)}{ϕ(4N)} ∑_{m=1}^{∞} \frac{a(m^{2})}{m^{2}} ∏_{p|4Nm^{2}}(1 + \frac{1}{p})^{−1} ∏_{p|4Nm^{2}}(1 − \frac{1}{p^{2}} )

and this is

\frac{1}{2} \frac{Y (log^{2} X)}{ϕ(4N)} ∑ \frac{a(m^{2})}{m^{2}} ∏_{p|4Nm^{2}}(1 − \frac{1}{p}).

Comparing this with the main term from the integral, we deduce that the total main term is

(\frac{1}{ϕ(4N)} ∑_{m=1}^{∞} \frac{a(m^{2})}{m^{2}} \frac{ϕ(4Nm)}{4Nm}) (\frac{1}{2}Y log^{2} X − \frac{1}{2}Y log^{2} \frac{X}{Y^{2}} ) + O(Y log X).

This is equal to

\frac{1}{2}CY (log Y^{2})(log \frac{X^{2}}{Y^{2}} ) = 2CY (log Y)(log \frac{X}{Y}).

§6. Concluding remarks.

Our theorem can be generalised to include the case when D ranges over an arithmetic progression mod M. This gives rise to a more complicated constant. It is also possible to prove Waldspurger’s theorem on the existence of a quadratic twist which does not vanish at s = 1 by these methods. Though we have not developed it here, the method allows one to obtain asymptotic formulas of the form

∑_{0<−D≤Y} L_{D}^{(j)}(1) = C_{j}Y log^{j} Y + O(Y log^{j−1} Y)

as Y →∞, for every j ≥ 1. The results of these investigations will appear in a future work.

Acknowledgements. We would like to thank David Rohrlich and Henryk Iwaniec for their comments on a preliminary version of this paper.

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