p-adic interpolation of square roots of central values

of Hecke L-functions

Adriana SoferMath Department, Princeton University

asofer@math.princeton.edu

March 24, 2000

Contents

1 Introduction 2

2 General setting 2

3 The reasonable modular forms θk− 1

2

(z) 5

3.1 Complex modular forms . . . . . . . . . . . . . . . . . . . . . 53.2 Operators on complex modular forms . . . . . . . . . . . . . . 73.3 Theta series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Complex integral-weight modular forms; homogeneous version 12

4 p-adic modular forms 164.1 Serre’s theory of p-adic modular forms . . . . . . . . . . . . . 174.2 Katz’s theory of p-adic modular forms . . . . . . . . . . . . . 174.3 Comparison of the theories and applications . . . . . . . . . . 19

5 p-adic interpolation of (8q d

dq

)rθ 1

2+δ/θ 1

2+δr≥0 25

6 Final p-adic interpolation 28

7 Appendix: Computation of the Euler Factor 37

1

1 Introduction

Let K = Q(√−l), l a prime, l ≡ 3 (mod 4). Let hK be its class number.

Let L(ψk, s) be the L-function corresponding to the Hecke character ψk =ψ2k−1 of K, where ψ is a fixed Hecke character of K of conductor

√−l whose

values on principal ideals lie in K.A theorem of Rodriguez-Villegas and Zagier ([22]) proves that the valuesL(ψk, k) are (up to the standard factors) squares in a fixed finite extensionof Q, and provides a formula, including the signs, for the square roots.Moreover, the formula is basically the evaluation of theta functions (or theirderivatives) at CM -points. Let

√L(ψk, k) be that square root (defined in

equation (5)).Let p > 3 be a rational prime that splits completely in K, and whose

prime factors in K are principal. Fix an isomorphism C ' Cp, the completionof an algebraic closure of Qp. We interpret C → Cp via this isomorphism.Every time we choose a prime ideal above p, we do it respecting the topologyin Cp. We are able to interpolate p-adically these “square roots” (that thiswas possible to do was conjectured by Koblitz in [15]). Our techniques followthe ideas of Katz ([12]) and Serre ([26]) of interpolating the modular formsinvolved and evaluating afterwards.

Our main result is:Theorem: If p > 3 does not divide hK , there is a continuous function Lp

on Zp with values in an integral extension of Zp such that:

Lp(k)

cp,k= (Euler Factor)

√L(ψk, k)

c∞,k

for a fixed class k ≡ k0 (mod 2hK(p − 1)), defined in theorem 6.17, wherecp,k is a power of a p-adic period, and c∞,k involves a power of a complexperiod and standard factors.Moreover, Lp is the Gamma-transform of a measure on Zp.

For the exact statement, see theorem 6.17, and use Lp(s) = Jδ(s−δ−1

2 ).

2 General setting

Let l > 3 , l ≡ 3 (mod 4) be a prime number, δ = 0, 1 with l+14 ≡ δ

(mod 2), k ≥ 1 be an integer, k ≡ δ + 1 (mod 2).Let K = Q(

√−l) ⊂ C, O = OK be the ring of integers of K, and CL(K) be

2

the class group of K. In this situation, the class group has odd order.

Let us extend ε(n) =(−ln

)=

(nl

), the Dirichlet character associated to K,

to OK by ε(µ) =(

2(µ+µ)l

).

Let ψk be a Hecke character of K satisfying:

ψk((α)) = ε(α)α2k−1 (1)

for α ∈ OK . It has conductor (√−l) . There are hK = |CL(K)| such

characters.Define the Hecke L-series

L(ψk, s) =∑

a

ψk(a)

Nas

where a runs over the non-zero integral ideals of K. It converges for Re(s) >k + 1

2 , but can be analytically continued to s ∈ C; it satisfies a functionalequation under s 7→ 2k − s with root number (−1)k+δ+1.

Let a run through a set of odd integral ideals, representatives of theideal class group of K, chosen such that a2 is primitive (not divisible by arational integer > 1). For each such a, it is not hard to see that we can write

a2 = Za2 + Z b+i√

l2 , where a = Na and b ∈ Z is defined mod 2a2; we choose b

in a fixed class (mod 16). We put za2 = b+i√

l2a2 .

Theorem 2.1 (Rodriguez-Villegas/Zagier)

L(ψk, k) =πkl

2k−3

4

2k−3(k − 1)!

∣∣∣∣∣∣

∑

[a]∈CL(K)

(−4

Na

)δ

ψk(a)−1θk− 1

2

(za2)

∣∣∣∣∣∣

2

(2)

where, for z = x + iy, y > 0,

θk− 1

2

(z) =1

(2πy)k−1

2

∑

n≥1,odd

(−1)n−1

2(k−1)Hk−1

(n

√πy

2

)eπin2 z

4 (3)

and the Hk’s are the Hermite polynomials defined in equation (7).

The expression(−4Na

)δψk(a)

−1θk− 1

2

(za2) depends only on ψk and [a] ∈ CL(K).

In particular it is independent of the choice of b.

3

Proof: See [22, Main Theorem], [21].

Let us rewrite the equality (2):

[eπi/42k−3(k − 1)!π−kl−

2k−3

4

]L(ψk, k) =

∑

[a]

(−4

Na

)δ

ψk(a)−1θk− 1

2

(za2)

2

(4)This provides us with a natural choice for the square root:

√L(ψk, k) = (5)

∑

[a]

(−4

Na

)δ

ψk(a)−1θk− 1

2

(za2)

e−πi/8

[2k−3(k − 1)!π−kl−

2k−3

4

]− 1

2

We want to interpolate p-adically the expression that appears squared on theright-hand side of equation (4), after adjusting some constants. Damerell’s

theorem ([1, 2]) shows that(

2πiΩ

)2k−1 [eπi/42k−3(k − 1)!π−kl−

2k−3

4

]L(ψk, k)

is algebraic for any appropriate period Ω. This motivates the next definition.We choose the branch of the square root function on C − r ∈ R, r ≤ 0

with positive real part, and extend it to C by defining r1

2 = i√−r. We

denote it by z1

2 . For k ∈ Z, z ∈ C∗, zk2 = (z

1

2 )k.

Definition 2.2 Let

la(k) =

(2πi

Ωa

)k− 1

2(−4

Na

)δ

ψk(a)−1θk−1/2(za2)

where Ωa = ζa2πi(−l)−1/4η2(za2) for a specific 12th root of unity ζa, a periodstudied by Gross in [7, 8], and η denotes the Dedekind eta function. Lemma6.4 shows that la(k) is algebraic.

We then want to interpolate the algebraic numbers

(2πi

ΩO

)k− 1

2 ∑

[a]

(Ωa

2πi

)k− 1

2

la(k).

We first work with la(k)k≥1 and obtain a partial interpolation.Remark: It follows from the functional equation that if k ≡ δ (mod 2),L(ψk, k) = 0. So we are really interested in the sequence la(2r+1+δ)r≥0.

4

Let us choose ψ1 as in (1), and ψk = ψ2k−11 for all k ≥ 1.

In [12], Katz p-adically interpolates the values of L(ψk, s) (a function of(k, s)) at (k, r), k, r ≥ 1 integer (removing the appropriate constants andEuler factors). Moreover, the resulting interpolation is the Gamma trans-form of a two-variable measure.

We will consider the values of the above L-function at the points of theanticyclotomic line, (k, k) (centers of symmetry for the functional equation).These are exactly the values studied in Theorem 2.1.

In [15], Conjecture B(i), Koblitz, following evidence that they are essen-tially squares of integers, conjectures that there should be a right choice ofsigns (independent of p) for the square roots that could be p-adically inter-polated (again removing the appropriate constants and Euler factors). Infact, Koblitz considers also twists of ψk, but, unfortunately, there is not aformula for these square roots yet.

Theorem 2.1 gives us a formula for the square root, including a sign, andthat permits us to prove the above conjecture for p > 3 not dividing hK .

3 The reasonable modular forms θk− 1

2

(z)

Notation: The symbol ∂∂h denotes the derivation in C characterized by

∂∂hh = 1, and ∂

∂h h = 0. Recall that we choose the branch of the squareroot function on C − r ∈ R, r ≤ 0 with positive real part, extended to C

by r1

2 = i√−r, and denoted by z

1

2 . For k ∈ Z, z ∈ C∗, zk2 = (z

1

2 )k. LetH = z ∈ C : Im(z) > 0.

3.1 Complex modular forms

This subsection follows the ideas found in G. Shimura’s work ([29, p.493]and [30, 31]). Also see [14, Ch.IV]. We present, without proofs, the basicfacts about Fourier expansions of half-integral weight modular forms.

If v : Γ −→ µR, and k ∈ 12Z, consider the set (Γ, v) = (γ, v(γ)), γ ∈ Γ.

It is a multiplier system for Γ. If (k, s) ∈ 12Z × C, F : H −→ C , (γ, v(γ))

(k, s)-acts on F by:

(F |(k,s) (γ, v(γ))

)(z) = F (γz)v(γ)−1(cz + d)−k | cz + d |−2s

Note: the index (k, 0) is often replaced by the index k.

5

Let (k, s) ∈ 12Z × C and let Γ be a congruence subgroup. The “r” in the

next definition stands for “reasonable”.

Definition 3.1 Let Mr(k,s)(Γ, v) be the set of functions F : H −→ C satis-

fying:

1. F is reasonable, that is, if z = x + iy, F can be expressed as a poly-nomial in the variables y and y−1, having meromorphic functions of zas coefficients,

2. F has the (k, s)-modular property for every (γ, v(γ)) ∈ (Γ, v); (k, s) iscalled the weight of F .

Lemma 3.2 Let F (z) =∑M

k=−M ykgk(z) be a reasonable function. Thenthe ordered set of meromorphic functions gk−M≤k≤M is uniquely deter-mined.

Proof: Left to the reader.

Let F : H −→ C be a holomorphic function having the (k, s)-modularproperty for a level N congruence subgroup Γ, (then, necessarily, s = 0),together with a multiplier system v with vR = 1. Let q = e2πiz, qM =e2πiz/M . For every cusp t, and every γ ∈ SL2(Z), γ.∞ = t, there is a Fourierexpansion of F :

Ft,γ(qRN ) = (Rth root of unity)∑

n∈Z

ct,γ(n)qnRN

for some ct,γ(n) ∈ C. (See, for example, [14, p.182]).

Definition 3.3 Let k ∈ 12Z. We call F holomorphic at the cusp t if

Ft,γ(qRN ) = (Rth root of unity)∑

n≥0

ct,γ(n)qnRN

for one (and therefore all) γ ∈ SL2(Z), γ.∞ = t.

Note that all our expansions are power series in qRN . It could happen that,at a particular cusp, RN might be replaced by some positive divisor. Thesmallest such integer is called the “width” of the cusp.

Let F : H −→ C be a function.

6

Definition 3.4 Let k ∈ 12Z. We say that F is a holomorphic modular form

of weight k with respect to (Γ, v) if

1. F is holomorphic on H,

2. F |k γ = F for every γ ∈ (Γ, v),

3. F is holomorphic at every cusp.

We denote Mk(Γ, v) the set of modular forms of weight k for (Γ, v).

Notation: For any ring A ⊂ C, define Mk(Γ, v, A) as the subset of Mk(Γ, v)consisting of the functions F such that Ft,γ ∈ A[[qNR]] for all t and γ. Suchan F is invariant under z 7→ z +RN . Fourier analysis tells us that there is aFourier expansion F (z) = F (qNR). We can always adjust the root of unityin order to have F∞,I = F .

3.2 Operators on complex modular forms

Definition 3.5 For F ∈ Mr(k,s)(Γ, v), k ∈ 1

2Z, s ∈ Z, define:

DF (z) =1

2πi

∂

∂zF (z) − k + s

12P (z)F (z),

WF (z) =1

2πi

∂

∂zF (z) − k + s

4πyF (z).

Remark: It is immediate to see that, if F is reasonable, then DF and WFare also reasonable.

The following is a general result.

Lemma 3.6 Let F and G be functions on H, F differentiable on H, k ∈ R.Define F ∗ by

F ∗(z) =1

2πi(

∂

∂zF )(z) − kG(z)F (z)

Then:1. if F (az+b

cz+d) = v(γ)(cz + d)kF (z)

for some γ =

(a bc d

)∈ SL2(Z), some v(γ) ∈ C,

and G(z) = 1(cz+d)2

G(az+bcz+d) − c

2πi(cz+d)

7

then

F ∗(az + b

cz + d) = v(γ)(cz + d)k+2F ∗(z)

2.F ∗ | [γ]k+2

F | [γ]k=

F ∗

F

Proof: easy computation.Examples of such G’s are:

• G(z) = 14πy ; F ∗(z) = (WF )(z)

• G(z) = 112P (z) = 1

12 − 2∑

n≥1 σ(n)e2πinz; F ∗(z) = (DF )(z) (Recallthe functional equation for P : z−2P (−1/z) = P (z) + 12

2πiz ).

Lemma 3.7 Let (k, s) ∈ 12Z × C. Let A ⊂ C be a subring.

1. If F ∈ Mr(k,s)(Γ, v), then DF, WF ∈ Mr

(k+2,s)(Γ, v),

2. If F ∈ Mk(Γ, v, A), then DF ∈ Mk+2(Γ, v, A[ 112 ]),

3. If F ∈ Mr(k,s)(Γ, v), G ∈ Mr

(l,t)(Γ, v), then D(FG) = DF.G + F.DG

and W (FG) = WF.G + F.WG.

Proof:1. and 2. are corollaries to lemma 3.6.3. follows from the definition.

Let

Lαk (z) =

k∑

j=0

Γ(k + α + 1)

Γ(j + α + 1)

(−z)j

j!(k − j)!(6)

be the general Laguerre polynomial, defined for α ∈ C.

Lemma 3.8 If F ∈ Mk(Γ, v), k ∈ 12Z, and F (qR) =

∑n≥0 anqn/R, an ∈ C,

q = e2πiz, then:

W rF (z) = (−1

4πy)rr!

∑

n≥0

Lk−1r (4πny/R)anqn/R

for r ≥ 1.

8

Sketch of proof: We proceed by induction on r, using lemma 3.7 (3) for

W rF =

wt=(0,r)︷ ︸︸ ︷(−1

4πy)rr!

wt=(k+2r,−r)︷ ︸︸ ︷∑

n

Lk−1r (4πny/R)anqn/R

︸ ︷︷ ︸wt=(k+2r,0)

and (for (2) and (3) see [19]):

1. Wy = 0,

2. x∂Lαn

∂x = nLαn(x) − (n + α)Lα

n−1(x),

3. (n + 1)Lαn+1(x) + (x − α − 2n − 1)Lα

n(x) + (n + α)Lαn−1(x) = 0.

Lemma 3.9 If F ∈ M 1

2Z(Γ, v) does not vanish on H, then

DF

F∈ M2(Γ)

Moreover, if A ⊂ C is a ring, F ∈ M 1

2Z(Γ, v, A), Γ of level N , vM = 1,

R = NM , F ∈ A[[qR]]∗, then

DF

F∈ M2(Γ, A[ 1

12R])

Proof: Notice that the quotient of two modular forms for (Γ, v) is modular

for Γ, since the quotient of the multiplier systems at γ =

(a bc d

)becomes

a power of cz + d. Since D increases the weight by 2, and F 6= 0 on H,DFF ∈ Mr

2(Γ).The only difficulty left is the possible zeroes of F at the cusps.Assume F (qR) =

∑n≥m anqn/R for some m ≥ 0, am 6= 0. Then

D(F )(qR) =∑

n≥0

(

n

R− k

12)an + 2k

∑

i+Rt=n

aiσ(t)

qn/R.

limy→∞

DF (qR)

F (qR)=

m

R− k

12

9

So if the order of vanishing of F at ∞ is m, then the order of vanishing ofDF at ∞ is at least m, and is exactly m whenever m

R − k12 6= 0.

By lemma 3.6,(DF )(z) | [γ]k+2r

F (z) | [γ]k=

(DF )(z)

F (z),

thus relating the behaviour at the other cusps with the one at ∞. HenceDFF ∈ M2(Γ).

If F ∈ A[[qR]]∗, then DF ∈ A[ 112R ][[qR]], and DF

F ∈ M2(Γ, A[ 112R ]).

Lemma 3.10 If F ∈ M 1

2Z(Γ, v) does not vanish on H, then

W rF

F∈ Mr

2r(Γ).

Proof: Same argument as before.

3.3 Theta series

A classical reference for the holomorphic theta-series is Weber ([36, chaptersII, III]).

Let

Hk(z) =

[ k2]∑

j=0

(−1)j k!

j!(k − 2j)!(2z)k−2j (7)

be the Hermite polynomial.Then we have:

Lemma 3.11 For k ≥ 0, the following equalities in Z[X]hold:

1. L−1/2k (X2)(−1)k22kk! = H2k(X)

2. L1/2k (X2)(−1)k22kk! = 1

2X H2k+1(X)

Proof: take α = −1/2, 1/2 in (6), and compare to (7).

The forms θk− 1

2

(z) were defined in equation (3). They are reasonable

functions, modular of weight k− 12 . For δ = 0, 1, θ2r+δ+ 1

2

∈ Mr2r+δ+ 1

2

(Γ0(2−δ), vδ) (see definition 3.1), where the description of vδ follows.

10

For γ =

(a bc d

)∈ Γ0(2), and d > 0, v0(γ) =

(cd

)e

πi4

(d−1+bd). Extend to

Γ0(2) by putting v0

(−1 00 −1

)= −i. Therefore,

θ2r+ 1

2

(az + b

cz + d) = v0(γ)(cz + d)2r+ 1

2 θ2r+ 1

2

(z) (8)

For γ =

(a bc d

)∈ SL2(Z), v1(γ) is defined by v1(S) = e

−3πi4 , v1(T ) = e

πi4 ,

v1

(−1 00 −1

)= i, where S =

(0 −11 0

)and T =

(1 10 1

).

If γ ∈ Γ0(2), and d > 0, v1(γ) = ( cd)e

πi4

(3(d−1)+d(b−c)−(d2−1)ac). Therefore,

θ2r+ 3

2

(az + b

cz + d) = v1(γ)(cz + d)2r+ 3

2 θ2r+ 3

2

(z) (9)

Notice that in both cases v(T 8) = 1, and v8 = 1.The first two,

θ 1

2

(z) = η2(2z)η(z) =

∑n≥1,odd eπin2z/4 ∈ M 1

2

(Γ0(2), v0)

θ 3

2

(z) = η3(z) =∑

n≥1,odd(−1)n−1

2 neπin2z/4 ∈ M 3

2

(SL2(Z), v1)

are holomorphic modular forms of weight 12 and 3

2 , respectively.Recall the product formula for the Dedekind eta function:

η(z) = eπiz/12∏

n≥1

(1 − e2πinz)

We immediately see that θ 1

2+δ has no zeroes in H.

The cusps for Γ0(2) are ∞, 0,−1, and the q-expansions are:

θ 1

2 ∞,I(q8) =

∑

n≥1,odd

qn2

8

θ 1

2 0,S(q8) = (8th root of unity)

1

2+

∑

n≥1

(−1)nq4n2

8

θ 1

2 −1,T−1S(q8) = (8th root of unity)

1

2+

∑

n≥1

q4n2

8

The remaining forms can be obtained from θ 1

2+δ using the Maass-Weil op-

erator.

11

Lemma 3.12 For r ≥ 0,

θ2r+δ+ 1

2

(z) = (8W )rθδ+ 1

2

(z) ∈ Mr2r+δ+ 1

2

(Γ0(2 − δ), vδ)

Proof: Let us prove the case δ = 0. The other case is similar. Equation(3) says

θ2r+ 1

2

(z) =1

(2πy)r

∑

n≥1,odd

H2r

(n

√πy

2

)eπin2 z

4 .

By lemma 3.11,

θ2r+ 1

2

(z) =1

(2πy)r

∑

n≥1,odd

(−1)r22rr!L−1/2r

(n2πy

2

)eπin2 z

4

=

(−2

πy

)r

r!∑

n≥1,odd

L−1/2r

(n2πy

2

)eπin2 z

4

= (8W )rθ 1

2

(z),

as seen in lemma 3.8 with R = 8, k = 1/2, the weight of θ1/2.We now apply lemma 3.7 (1) to deduce that θ2r+ 1

2

∈ Mr2r+ 1

2

(Γ0(2), v0).

Corollary: Since θ 1

2+δ has no zeroes in H, we see that, moreover:

θ2r+δ+ 1

2

θ 1

2+δ

∈ Mr2r(Γ0(2 − δ))

Proof: use lemma 3.10.

3.4 Complex integral-weight modular forms; homogeneous

version

We will relate integral weight modular forms to p-adic modular forms. Forthis, we introduce a homogeneous version of reasonable modular forms ofweight (k, s) ∈ Z × C (see lemma 3.17 below).

Let M = (ω1, ω2) ∈ C∗2 : Im(ω1

ω2) > 0. We can think of M as the set

of oriented basis of lattices in C.In this context, we say that f is a modular form for a congruence subgroup

Γ if f(ω′1, ω

′2) = f(ω1, ω2) for every γ ∈ Γ, where γ

(ω1

ω2

)=

(ω′

1

ω′2

).

We say that f has weight (k, s) ∈ Z × C if, for every λ ∈ C∗, f(λω1, λω2) =λ−k | λ |−2s f(ω1, ω2).

12

One example is the area function A(ω1, ω2) = − 14π2 Im(ω1ω2). We have

A(2πiz, 2πi) = y ∈ Mr(0,−1)(SL2(Z)), where z = x + iy ∈ H.

We can easily check that functions F ∈ Mrk(Γ), k ∈ Z, give rise to

modular forms f of weight k in this sense, by defining f : M −→ C,

f(ω1, ω2) =

(2πi

ω2

)k

F (ω1/ω2) (10)

If f : M −→ C is modular for SL2(Z), then the values of f at every(ω1, ω2) ∈ M generating a given lattice L ⊂ C coincide. Consider E anelliptic curve over C isomorphic to C/L. To recover L from E, we need tofix a nowhere vanishing invariant differential ω. We say that a pair (E, ω) isisomorphic to (E′, ω′) if there is an isomorphism of elliptic curves h : E −→E′ such that ω = h∗ω′. Notice that in the case (C/L, ω) ' (C/L′, ω′) thereexists λ ∈ C∗ with L′ = λL and ω′ = λ−1ω.

The correspondence between lattices L in C and isomorphism classes ofpairs (E, ω) of elliptic curves with nowhere vanishing differentials:

L 7−→ (E, ω) = (C/L, dz)

(where z is the variable in C) is one to one, the inverse being:

(E, ω) 7−→ L =

∫

Cω, C loop in E

.

L is called the lattice of periods of (E, ω). The space of invariant differen-tials over E has dimension 1 over C, so ω = Ωdz for some Ω ∈ C∗. Changingω by λω for some λ ∈ C∗ induces a change of L by λL. We can considerany f, modular for SL2(Z), and of weight k ∈ Z, as a function of isomor-phism classes of pairs (E, ω) of elliptic curves over C together with nowherevanishing differentials over E (called test objects). We denote it again by f.

Definition 3.13 Take L ↔ (E, ω). Write L = ω1Z+ω2Z for some (ω1, ω2)∈ M .

f(E, ω) = f(C/L, dz) = f(ω1, ω2)

Good definition: we check that if (ω′1, ω

′2) is another element of M gener-

ating the same lattice as (ω1, ω2), the change of basis matrix is in SL2(Z).Therefore f(ω′

1, ω′2) = f(ω1, ω2).

Lemma 3.14 for λ ∈ C∗, f(E, λω) = λ−kf(E, ω).

13

Proof: Left to the reader.Remark: According to equation (10) and the previous discussion, everyF ∈ Mr

k(SL2(Z)), k ∈ Z, gives rise to f, a function of test objects (E, ω).

We are going to work with level N congruence subgroups Γ, not onlywith SL2(Z). We introduce some technical modifications to the previousdiscussion. A definition like 3.13 would only be correct if the change ofbasis matrix from (ω1, ω2) to (ω′

1, ω′2) is in Γ. So we modify our test objects

(E, ω), adapting the definition to this situation.

Definition 3.15 Given an elliptic curve E, we call a Γ(N)-structure on Ean isomorphism of groups α : Z/NZ × Z/NZ −→ NE , where NE is thesubgroup of elements of order dividing N in E.

Note: if E ' C/L, then NE ' 1N L/L.

Definition 3.16 A Γ(N)-test object is a triplet (E, ω, α). We say that twoobjects are equivalent, and denote it (E, ω, α) ' (E′, ω′, α′) if there is anisomorphism h : E → E′ with ω = h∗ω′, and α = h∗α′.

We can consider every f, modular for Γ(N), of weight k ∈ Z, as a functionof equivalence classes of test objects (E, ω, α). We denote it again by f. Tofind the evaluation f(E, ω, α), we have to

• find the lattice L ⊂ C such that (E, ω) ↔ L

• find α′ with α = h∗α′, if h : E −→ C/L is the corresponding isomor-phism. Then (E, ω, α) ' (C/L, dz, α′) (for example, (C/Ω−1L,Ωdz, α)// '(C/L, dz, Ωα)).

• find a basis (ω1, ω2) ∈ M of L such that α′(x, y) = xω1

N +y ω2

N (mod L).

Definition 3.17

f(E, ω, α) = f(C/L, dz, α′) = f(ω1, ω2)

This is well defined (the evaluation is independent of the choice of basis).Note: We do not have a notion of cusp here, but the different level Nstructures provide the equivalent information.Remark: We can safely work with Γ(N)-test objects (that is, with def-initions 3.15, 3.16 and 3.17), even for modular forms for congruence sub-groups Γ ⊃ Γ(N). Note that if (E, ω) ↔ L = Zω1 + Zω2, αM

NM = αN , and

14

α′N (x, y) = xω1

N +y ω2

N (mod L), then α′NM (x, y) = x ω1

NM +y ω2

NM (mod L).If f is modular for Γ(N), then f(E, ω, αNM ) = f(E, ω, αN ). However, for somecongruence subgroups, we can loosen up the definition of level N structure.For Γ1(N), we take a level N structure to be an inclusion α : Z/NZ → NE.The evaluation given in definition 3.17 is adapted by asking that (ω1, ω2) ∈M satisfy α′(y) = y ω2

N (mod L).For Γ0(N), we take a level N structure to be the image of an inclusionα = Im(Z/NZ → NE). The evaluation given in definition 3.17 is adaptedby asking that (ω1, ω2) ∈ M satisfy that α′ be generated by ω2

N (mod L).For more details on level structures see [12, p.477], [14, p.153, “modular

points”].

Lemma 3.18 In any of the above cases, if f is modular for Γ of weightk ∈ Z, and λ ∈ C∗, then

f(E, λω, α) = λ−kf(E, ω, α).

Proof: Left to the reader.Note: Equation (10) and the previous discussion show that every F ∈Mr

k(Γ), k ∈ Z, gives rise to f, a function of test objects (E, ω, α).

Let f be modular for Γ(N). For τ ∈ H, let Lτ = 2πiτZ+2πiZ. Let ατ bethe level N structure ατ (x, y) = x2πiτ

N + y 2πiN (mod Lτ ) and qN = e2πiτ/N .

Definition 3.19f(qN ) = f(C/Lτ , dz, ατ )

(we see that the right-hand side is invariant under τ 7−→ τ + N , since

TN =

(1 N0 1

)∈ Γ(N)).

We call f(qN ) the qN -expansion of f.

If F ∈ Mrk(Γ(N)), k ∈ Z, and f is the associated function of test objects,

then

1. F (qN ) = f(qN ),

2.Ft,γ(qN )

t cusps,γ.∞=t

= f(C/Lτ , dz, α)α all level N structures.

We will often abuse the notation, and use F for f.

15

Remark: The exponential map induces the isomorphism of test objects

(C/Lτ , dz, ατ ) ' (C∗/qZ, dt/t, eατ ),

where q = e2πiτ , t is the variable in C∗, and eατ (x, y) = qxN e

2πiyN (mod qZ).

Recall the isomorphism

C/Lh−→ (X, Y ) ∈ C2 : Y 2 = 4X3 − g2(L)X − g3(L) ∪ ∞

z (mod L) 7−→ (X = ℘(z), Y = ℘′(z))

0 (mod L) 7−→ ∞ (11)

and dz = h∗ dXY , where g2(L) = 60

∑′l∈L l−4 ∈ M4(SL2(Z)) and g3(L) =

140∑′

l∈L l−6 ∈ M6(SL2(Z)), and ℘ is the Weierstrass function.

Definition 3.20 If h∗(α∗τ ) = ατ , we interpret

(C/Lτ , dz, ατ )h' (Y 2 = 4X3 − g2(L)X − g3(L),

dX

Y, α∗

τ )

as a test object over Z[16 ]((qN )), and denote it (Tate(q), ωcan, αcan), the Tatecurve together with the canonical differential, and canonical level N struc-ture. ([9, A1.2], [32, App.C.14]).

Observation: The interpretation of the q-expansion as evaluation at theTate Curve gives a purely algebraic approach that does not need Fourieranalysis, and that can be used in other (non-complex) situations.

4 p-adic modular forms

Let p be an odd prime. Assume p ≥ 5 in order not to worry about 112 .

Let L be a number field, OL its ring of integers, L℘ its completion at aprime ℘ in L above p, O℘ the ring of integers of L℘, L℘ the completion ofthe maximal unramified extension of L℘, O℘ its ring of integers.Let B be a p-adic ring (i.e. B = limn B/pnB).

There are two theories of p-adic modular forms of (essentially) integralweight: Serre’s and Katz’s. We briefly present and compare them.

16

4.1 Serre’s theory of p-adic modular forms

Let X be the group of continuous p-adic characters of Z∗p. X is isomorphic

to Zp×Z/(p−1)Z, by viewing its elements as (t, u), where z →< z >t ω(z)u.We may view Z as a dense subgroup of X (where n ∈ Z corresponds to thecharacter z → zn).Let Γ be a level N congruence subgroup, and v a multiplier system on Γ.Let M ≥ 1 be an integer such that vM = 1 (see subsection 3.1 for notation).Let R = NM .

Definition 4.1 (Serre) We call a p-adic modular form over B for (Γ, v),a collection Ft(qR)t cusp ⊂ B[[qR]] such that there is a sequence ofFj ∈ Mkj (Γ, v, B ∩ C) whose q-expansions

Fj,t −→ Ft

uniformly on the coefficients, for every cusp t. The limit k = lim kj ∈ Xexists; it is called the weight of f . (See note below). For k ∈ X, we denoteMp

k(Γ, v, B) the set of such forms.

Note: Existence of the weight. See [26, theorem 2]) and [9, corollary 4.4.2]for the case v = 1. If vM = 1, take θ ∈ Mm(Γ, v, B) for some m ∈ Z

and apply the v = 1 case to FjθM−1 ⊂ Mkj+m(M−1)(Γ, B). We get

kj + m(M − 1) converges to a limit in X, and so does kj too.

4.2 Katz’s theory of p-adic modular forms

Katz chooses a modular approach; i.e. his generalized p-adic modular formsare defined as functions of equivalence classes of p-adic test objects.

Definition 4.2 Let B be a p-adic ring. Let E be an elliptic curve over B.We call ϕ a trivialization of E/B if ϕ : E −→ Gm is an isomorphism overB, where E is the formal group associated with E ([32], [18, Appendix 1.3by J. Tate]), and Gm is the multiplicative formal group over B.We call α a level N structure on E/B if α : Z/NZ × Z/NZ −→ NE isan isomorphism of group schemes over B (NE is the group of B-points oforder dividing N) under which the canonical pairing eN :N E ×N E −→ µN

becomes < (η1, n), (η2, m) >= ηm1 /ηn

2 .A trivialization and a level N structure are compatible if, writing N = prN0,p 6 |N0, µpr → E

ϕ−→ Gm is the inclusion.

17

(compare to the complex level N structures in subsection 3.4).A triplet (E, ϕ, α) with ϕ and α compatible is called a p-adic level N testobject over B.

Note: The previous definition of level N structure is arithmetic. There isalso a “naave” level N structure, α : Z/NZ×Z/NZ −→ NE, an isomorphismof group schemes over B. When 1/N ∈ B and µN ⊂ B, the notions of naiveand arithmetic level N structures coincide. We only need to work with alevel N divisible by p (that is 1/N 6∈ B), and only very briefly, when usingthe “Frobenius” map in the Appendix.

We need E to be of ordinary reduction, and to have a lot of constants tohave a trivialization. If E has ordinary reduction and B ⊂ Cp is a completedvr whose residue field is algebraically closed, we have trivializations ([20,Corollary 4.3.3]). They all differ by a constant in Z∗

p = Aut(Gm). To readmore about formal groups, see [5] and [32, Chapter IV].

We defined (Tate(q), ωcan, αcan) as a test object over Z[16 ]((qN )) (see

definition 3.20). Since p ≥ 5, it is defined over C = Zp((qN )). There is a

canonical C-isomorphism T ate(q)ϕ7−→ Gm such that ϕ∗( dT

1+T ) = ωcan. Wecall it ϕcan ([12, 5.2]).

Definition 4.3 The level N test object (Tate(q), ϕcan, αcan) over Zp((qN ))is called the Tate curve together with canonical trivialization and canonicallevel N structure.

Let A ⊂ A′ ⊂ Cp be p-adic rings containing B.

Definition 4.4 (Katz) Consider the functions F of A-equivalence classesof p-adic level N test objects over A with values in A, commuting with ex-tension of scalars A → A′, and satisfying F(Tate(q), ϕcan, α) ∈ B[[qN ]] forevery level N structure over B. Define KMp(Γ(N), B) to be the set of suchfunctions. It is a p-adic ring ([12, page 504]).Let k ∈ X. Let KMp

k(Γ(N), B) be the subset of functions F that, in addi-tion, satisfy for λ ∈ Z∗

p, F(E, λϕ, α) = k(λ)F(E, ϕ, α). For more generaland detailed definitions, see [9, 10, 12].

Theorem 4.5 With notation as above, but (N, p) = 1, k ∈ Z, KMpk(Γ(N), B)

coincides with the set of functions F of A-equivalence classes of level Ntest objects (E, ω, α) over A with values in A, commuting with extension of

18

scalars A → A′, and satisfying F(Tate(q), ωcan, α) ∈ B[[qN ]] for every levelN structure over B, F(E, λω, α) = λ−kF(E, ω, α) for every λ ∈ A∗.

Proof: See [6, theorem I.3.5].

Definition 4.6 If F ∈ KMp(Γ(N), B), the value F(Tate(q), ϕcan, αcan) ∈B[[qN ]] is called the qN -expansion of F . The set of qN -expansions isF(Tate(q), ϕ, α) for all trivializations ϕ and all level N structures α.

Theorem 4.7 (q-expansion principle) Assume that (N ; p) = 1. Let F ∈KMp(Γ(N), B). Let µN ⊂ B′ ⊂ B be a p-adic subring. If there is one q-expansion of F with coefficients in B′, then F ∈ KMp(Γ(N), B′).

The q-expansion map is injective, and B((qN ))/KMp(Γ(N), B) is B flat.

Proof: [12, 5.2], [6, I.3.2].

Theorem 4.8 KMp(Γ(prN), B) ' KMp(Γ(N), B).

Proof: [12, lemma 5.6.3].

Classical modular forms of integral weight and coefficients in a p-aic ringB can be interpreted as p-adic modular forms.

4.3 Comparison of the theories and applications

In the presence of a weight k ∈ X, the two theories coincide:

Theorem 4.9 Assume that (N ; p) = 1. Let k ∈ X.Then Mp

k(Γ(N), B) = KMpk(Γ(N), B).

In particular, if B = O℘, (E, ω, α) is a (complex) level N test object overOL, F ∈ KMp

k(Γ(N), B), we can define F(E, ω, α) = limFkj (E, ω, α), Fk ∈Mkj (Γ(N),OL) a particular sequence such that Fkj −→ F (see [6, I.2.6]).

Proof: [6, I.3.4], [11, A.1.6].Note that (almost) every time we evaluate a form, we are invoking theprevious theorem.

Theorem 4.10 Assume that (N ; p) = 1, p ≥ 5. Let x ∈ B divide a powerof p. Let F ∈ KMp

k(Γ(N), B), k ∈ Z, k ≥ 0. If k = 1, we add the restrictionN < 12. The following are equivalent:

(i) F is congruent to 0 mod x as a form,(ii) F(qN ) ∈ xB[[qN ]],(iii) all q-expansions of F are in xB[[qN ]].

19

Proof: ([6, I.2.10], [9, 2.7.1 and 2.8.3]).

Corollary: The theorem shows that, under good conditions, convergenceat one cusp implies convergence at every cusp. We deduce that when k ∈ Z,definition 4.1 can be loosened up by asking only Fj,∞ −→ F∞. In this case,the p-adic modular form Ftt cusp is simply denoted F = F∞.

Forms in KMp(Γ(N), B) considered in its full generality, can only, inprinciple, be evaluated at test objects (E, ϕ, α). Given (E, ϕ, α), a p-adiclevel N test object over B, we obtain (E, ω, α), a level N test object overB, by putting ω=(the unique extension from E to E of) ϕ∗( dT

1+T ). Recall

that ϕ∗can( dT

1+T ) = ωcan.

Definition 4.11 With notation as above, if F ∈ KMp(Γ(N), B), defineF(E, ω, α) = F(E, ϕ, α).

Given a level N test object (E, ω, α) over B, we can only obtain a p-adiclevel N object (E, ϕ, α) over B such that ω = ϕ∗( dT

1+T ) if ω is “magic”(see[12, p.506]).

Theorem 4.12 If E/O℘ is ordinary, ω/E a differential, then there is c ∈O∗

℘, such that cω is magic.

Denote σ the Frobenius automorphism of L℘/L℘. Then cσ/c is the p-adicunit eigenvalue of Frobenius relative to O℘/℘ on E ⊗ O℘/℘. If O℘ = Zp,and c satisfies the previous equality, then cω is magic.

Proof: [4], [12, 5.4], [20, Corollary 4.3.3].

Let Z be the Z[ 112 ]-algebra generated by the operators: g2, g3, S, H, W, D.

Define a (complex) action on Mr•(Γ0(2 − δ)), the Z-graded ring, by:

• g2, g3 act by multiplication,

• S(z) = P (z) − 12 14πy ∈ M2(SL2(Z)) acts by multiplication, where

P (z) = 1 − 24∑

n≥1 σ(n)e2πinz is the Ramanujan P -function, σ(n) =∑d|n,d>0 d,

• H(f) = kf if f has weight k ∈ 12Z, and is additive,

• W is the Maass-Weil operator from definition 3.5, extended by addi-tivity,

20

• D = W − 112SH is the Halphen-Fricke operator from definition 3.5,

extended by additivity.

Z sends Mr•(Γ0(2 − δ), Z[ 1

12 ]) to itself, since W does so, H obviously doesso too, and g2, g3, S ∈ Mr

Z(SL2(Z)).

The (p-adic) action of Z on KMp(Γ0(2 − δ)), denoted ?, is:

• g2, g3 act by multiplication,

• S acts by multiplication by P (q),

• H acts as before,

• W acts as q ddq ,

• D = q ddq − 1

12 P (q)H.

Note that the complex and p-adic actions of S, W , and D differ; but we stillhave the relation D = W − 1

12SH.

Important Fact: P ∈ Mp2(SL2(Z)) (although P 6∈ MZ(SL2(Z))).

Proof (Serre): see [9, 4.5.4].

Lemma 4.13 q ddq is a weight 2 derivation on Mp

Z(Γ(N), Zp).

Proof: D and H send MZ(Γ(N), Z) into MZ(Γ(N), Z[ 112 ]). If F =

∑n≥0 anqn

8

has weight k, then

D(F) =∑

n≥0

(

n

8− k

12)an + 2k

∑

i+Rt=n

aiσ(t)

qn

8

This shows that the action of D is continuous. Therefore, D and H sendalso Mp

Z(Γ(N), Zp)) into itself.

As seen above, P ∈ Mp2(SL2(Z)). Write q d

dq = D + 112PH to end the proof.

Corollary: If Fθ 1

2+δ

∈ MZ(Γ(N),OL), N even, then,

(q ddq )rF

θ 1

2+δ

∈ MpZ(Γ(N),O℘) (12)

21

Proof:(q d

dq )rF

θ 1

2+δ

=

q

d

dq+

q ddq θ 1

2+δ

θ 1

2+δ

r F

θ 1

2+δ

To prove thatq d

dqθ 1

2+δ

θ 12+δ

∈ Mp2(Γ0(2 − δ), Zp), use

q ddq θ 1

2+δ

θ 1

2+δ

=

Dθ 1

2+δ

θ 1

2+δ

+

1

24P ,

whereDθ 1

2+δ

θ 12+δ

∈ M2(Γ0(2 − δ), Z[ 112 ]) (lemma 3.9).

We denote by CM(E) the field of complex multiplications of a CM-elliptic curve E.

Theorem 4.14 (Katz) Let F ∈ MZ(Γ,OL) (then F ∈ MpZ(Γ,O℘)). Con-

sider any level N test object (E, ω, α) defined over OL, L ⊃ CM(E), whereE is a CM-elliptic curve with good reduction at ℘, and p splits completelyin CM(E) (this is equivalent to E having ordinary reduction at ℘). Then

1. For every z ∈ Z, zF (E, ω, α) and z ? F (E, ω, α) lie in OL

2. zF (E, ω, α) = z ? F (E, ω, α)

Proof: [12, page 532].

Corollary: In particular, W r(F )(E, ω, α) = (q ddq )r(F )(E, ω, α).

We are going to prove a similar result for half-integral weight forms. Aclassical result states that a modular form of weight k ∈ 1

2Z for Γ0(4) and

multiplier system the kth power of that of θ(z) =∑

n∈Ze2πin2z (the standard

theta function) is a polynomial in θ and F , having pure degree equal to k,where θ has weight and degree 1

2 , F =∑

n≥1,odd σ(n)e2πinz = θ41

2

(2z) , has

weight and degree 2. This is treated, for instance in [14, chapter IV].We have similar results for θ 1

2+δ (see subsection 3.3):

Lemma 4.15 1. M 1

2Z(Γ0(2), v0) = MZ(Γ0(2), v0) ∪ θ 1

2

MZ(Γ0(2)).

22

2. M 1

2Z(SL2(Z), v1) = MZ(SL2(Z), v1) ∪ θ 3

2

MZ(SL2(Z)).

Proof:1. Assume f ∈ M 1

2Z(Γ0(2), v0) − MZ(Γ0(2), v0).

Then f(2z)θ71

2

(2z) ∈ MZ(Γ0(4)). By the previous result, there is a polyno-

mial P with f(2z)θ71

2

(2z) = P (θ4(z), F (z)). We look at the order of zero

at z = i∞ at the q-expansions and deduce that F 2 divides P (θ4(z), F (z)).Use the identity F (z) = θ4

1

2

(2z), and get f(z) = θ 1

2

(z)Q(θ4(z/2), θ41

2

(z)) for

some other polynomial Q. Finally, Q(θ4(z/2), θ41

2

(z)) ∈ MZ(Γ0(2)) since

θ(z/2) ∈ M 1

2

(Γ0(2), v0).

2. Assume f ∈ M 1

2Z(SL2(Z), v1)−MZ(SL2(Z), v1). Then, using the fact

that, for k ∈ Z, Mk(SL2(Z)) = CEk ⊕ ∆Mk−12(SL2(Z)), Mk(SL2(Z)) =0 if k < 0 or k ≥ 1 is odd, we get f(z)θ7

3

2

(z) = P (∆, E′js), for some

polynomial P . Again, we look at orders of vanishing at i∞, and deducethat ∆ divides P (∆, E′

js). Recall that ∆ = η24 = θ83

2

. Hence f(z) =

θ 3

2

(z)Q(g2, g3), for some polynomial Q. Finally, Q(g2, g3) ∈ MZ(SL2(Z))

since θ 3

2

∈ M 3

2

(SL2(Z), v1).

Let (E, ω, α) be as in theorem 4.14.

Theorem 4.16 Let F ∈ MZ+ 1

2

(Γ0(2 − δ), vδ,OL) (as seen in lemma 4.15,

this implies that Fθ 1

2+δ

∈ MZ(Γ0(2 − δ),OL)). Then

W r(F )

θ 1

2+δ

(E, ω, α) =(q d

dq )r(F )

θ 1

2+δ

(E, ω, α)

Proof: The fact that(q d

dq)r(F )

θ 12+δ

∈ MpZ(Γ0(2−δ),O℘) was proved in a previous

Corollary (cf. equation (12)). Write F = Gθ 1

2+δ, G ∈ MZ(Γ0(2 − δ),OL),

G ∈ MpZ(Γ0(2 − δ),O℘).

Define W ′(G) = 1θ 1

2+δ

W (Gθ 1

2+δ), an operator on MZ(Γ0(2 − δ)) into

MrZ(Γ0(2 − δ)).

Then:W r(F )

θ 1

2+δ

= (W ′)r(G)

23

Claim: W ′ = W +Wθ 1

2+δ

θ 12+δ

∈ Z.

Proof: W ′(G) = 1θ 1

2+δ

W (Gθ 1

2+δ) = WG + G

Wθ 12+δ

θ 12+δ

.

Wθ 12+δ

θ 12+δ

=Dθ 1

2+δ

θ 12+δ

+ 124S ∈ Z. (Recall that

Dθ 12+δ

θ 12+δ

∈ M2(Γ0(2−δ), Z[ 112 ]) ⊂ Z).

Hence,

W r(F )

θ 1

2+δ

= (W ′)r(G) = (W +Wθ 1

2+δ

θ 1

2+δ

)rG (13)

Claim:Wθ 1

2+δ

θ 1

2+δ

(E, ω, α) =q d

dq θ 1

2+δ

θ 1

2+δ

(E, ω, α)

Proof: Since W is a derivation,

Wθ 1

2+δ

θ 1

2+δ

= 1/8W (θ8

1

2+δ

)

θ81

2+δ

We are now dealing with forms of integral weight, and, since v8 = 1, trivialmultiplier. By theorem 4.14,

θ81

2+δ

(E, ω, α) = θ81

2+δ

(E, ω, α)

W (θ81

2+δ

)(E, ω, α) = qd

dq(θ8

1

2+δ

)(E, ω, α)

Therefore

1/8W (θ8

1

2+δ

)

θ81

2+δ

(E, ω, α) = 1/8q d

dq (θ81

2+δ

)

θ81

2+δ

(E, ω, α)

We finish by noting that

1/8q d

dq (θ81

2+δ

)

θ81

2+δ

=q d

dq θ 1

2+δ

θ 1

2+δ

∈ Mp2(Γ0(2 − δ), Zp)

(we are using that q ddq is a derivation, and the last Corollary).

So the last term in equation (13) corresponds to:

(qd

dq+

q ddq θ 1

2+δ

θ 1

2+δ

)rG =(q d

dq )rGθ 1

2+δ

θ 1

2+δ

=(q d

dq )rF

θ 1

2+δ

24

5 p-adic interpolation of (8q d

dq

)rθ 1

2+δ/θ 1

2+δr≥0

General facts about measures on Zp with values on a p-adic ring can befound in [12, p.520], [38, 33], [17, Ch.4].

By lemma 3.12, we have θ2r+ 1

2+δ = ˜(8W )rθ 1

2+δ. Now, by theorem 4.16,

(8W )rθ 1

2+δ

θ 1

2+δ

(E, ω, α) =(8q d

dq )rθ 1

2+δ

θ 1

2+δ

(E, ω, α) (14)

for every test object as described there.We compute

(8q ddq )rθ 1

2+δ(q8)

θ 1

2+δ

=

∑n≥1,odd

[(−1)

n−1

2 n]δ

n2rqn2

8

θ 1

2+δ

∈ Mp2r(Γ0(2 − δ), Zp)

We are going to define a function Iδ as the Gamma-transform of a measure

µ∗δ . This will provide us with the interpolation of

(8q d

dq

)rθ 1

2+δ

θ 12+δ

r≥0 (see

lemma 5.5). It is not hard to see that the Gamma-transform of a measure µis related to its r-moments,

∫Zp

xrdµ. The next definition is aimed at having

(8q d

dq

)rθ 1

2+δ

θ 12+δ

r≥0 as the moments of a measure.

Definition 5.1 Let µδ be the measure on Zp with values on the p-adic ringKMp(Γ(2 − δ), Zp) whose Fourier transform is:

µδ(X) =∑

n≥1,odd

[(−1)

n−1

2 n]δ

qn2

8

θ 1

2+δ

(X + 1)n2 ∈ (Zp[[q8]])[[X]]

Good definition:

µδ(X) =∑

m≥0

∑n≥1,odd

(n2

m

)[(−1)

n−1

2 n]δqn2

8

θ 1

2+δ(q8)

Xm

These coefficients belong to Zp[[q8]]. Put X = T − 1. By abuse of notation,we will use

25

µδ(q8, T ) = µδ(X) =∑

n≥1,odd

[(−1)n−1

2 n]δqn2

8

θ 1

2+δ(q8)

Tn2

The r-moments of µδ are

(Td

dT)rµδ |T=1 =

∫

Zp

xrdµδ(T )

=

∑n≥1,odd[(−1)

n−1

2 n]δn2rqn2

8

θ 1

2+δ(q8)

=(8q d

dq )rθ 1

2+δ

θ 1

2+δ

∈ Mp2r(Γ0(2 − δ), Zp)

The coefficients of µδ(X),∫

Zp

( xm

)dµδ(x), are a Q-linear combination of the

moments, so they belong to KMp(Γ(2−δ), Zp)⊗Qp∩Zp[[q8]]. Since the level2 − δ is coprime to p, we can apply the q-expansion principle (theorem 4.7)to an element of KMp(Γ(2− δ), Zp)⊗Qp ∩Zp[[q8]], that is included in the p-

torsion of(

Zp((q8))/KMp(Γ(2 − δ), Zp))

and deduce that they really are in

KMp(Γ(2−δ), Zp) (that p-torsion is trivial, since Zp((q8))/KMp(Γ(2−δ), Zp)is Zp-flat).

Lemma 5.2 The measure µδ|pZp, and extended by 0, on Zp with values inKMp(Γ(2 − δ), Zp), satisfy

µδ(q8, T ) |pZp= [(−1)p−1

2 p]δ∑

n≥1,odd

[(−1)n−1

2 n]δqp2n2

8

θ 1

2+δ(q8)

T p2n2

Proof:n ≥ 1, odd, n2/8 ≡ 0 (mod p) ⇔ n ≥ 1, odd, n ≡ 0 (mod p)⇔ n = pm, m ≥ 1, odd.Therefore

µδ(q8, T ) |pZp =1

θ 1

2+δ(q8)

∑

m≥1,odd

[(−1)pm−1

2 pm]δqp2m2

8 T p2m2

=1

θ 1

2+δ(q8)

∑

m≥1,odd

[(−1)p−1

2m(−1)

m−1

2 pm]δqp2m2

8 T p2m2

= [(−1)p−1

2 p]δ∑

n≥1,odd

[(−1)n−1

2 n]δqp2n2

8

θ 1

2+δ(q8)

T p2n2

26

Lemma 5.3 The moments of µδ|pZp are

[(−1)p−1

2 p]δp2r

[(8q d

dq )rθ 1

2+δ

](qp2

)

θ 1

2+δ(q8)

∈ KMp2r(Γ(2 − δ), Zp)

Proof:

(Td

dT)rµδ|pZp(q8, T ) |T=1=

[(−1)p−1

2 p]δ

θ 1

2+δ(q8)

∑

n≥1,odd

[(−1)

n−1

2 n]δ

(pn)2rqp2n2

8

=[(−1)

p−1

2 p]δp2r

θ 1

2+δ(q8)

∑

n≥1,odd

[(−1)

n−1

2 n]δ

n2rqn2

8

(qp2

)

= [(−1)p−1

2 p]δp2r

[(8q d

dq )rθ 1

2+δ

](qp2

)

θ 1

2+δ(q8)

They belong to Mp2r(Γ0(p

2(2− δ), Zp[µp2 ])∩KMp(gaol, Zp) = KMp2r(Γ(2−

δ), Zp) (see the discussion in [6, p.22]; the weight remains unchanged whenreducing the level because the nbentypus is trivial).

Definition 5.4 Let Iδ(q8, s) : Zp −→ KMp(Γ(2 − δ), Zp) be the Gamma-transform of the measure µ∗

δ, defined as follows:

Iδ(q8, s) =

∫

Zp

< x >s dµ∗δ(q8, T )

where µ∗δ = µδ |Z∗

p= µδ − µδ |pZp.

Let r → s ∈ Zp, | r |→ ∞, r ≡ 0 (mod p − 1). Then,

∫

Zp

xrdµδ(q8, T ) −→ Iδ(q8, s)

On the left-hand side, we have the r-moments of µδ; they belong toKMp

2r(Γ(2 − δ), Zp). We get that Iδ(q, s) ∈ KMp2s(Γ(2 − δ), Zp). Iδ(q, s)

is the Gamma-transform of the measure µ∗δ , therefore continuous in the

variable s.

27

Lemma 5.5 For r ≡ 0 (mod p − 1),

Iδ(q8, r) =(8q d

dq )rθ 1

2+δ(q)

θ 1

2+δ(q)

− [(−1)p−1

2 p]δp2r

[(8q d

dq )rθ 1

2+δ

](qp2

)

θ 1

2+δ(q)

Proof:

Iδ(q8, r) =

∫

Zp

< x >r d(µδ(q8, T ) − µδ |pZp (q8, T ))

For r ≡ 0 (mod p − 1),

Iδ(q8, r) =

∫

Zp

< x >r d(µδ(q8, T ) − µδ |pZp (q8, T ))

= (r-moment of µδ)- (r-moment of µδ |pZp)

=(8q d

dq )rθ 1

2+δ

θ 1

2+δ

−[(−1)

p−1

2 p]δ

p2r

[(8q d

dq )rθ 1

2+δ

](qp2

)

θ 1

2+δ

by lemma 5.3

Remark: For r ≡ r0 (mod p− 1), Iδ(q8, r) has a similar expression to theone in lemma 5.5, replacing θ 1

2+δ by the twists

θ 1

2,ω−r0 =

∑n≥1,odd ω−r0(n2/8)qn2

8

θ 3

2,ω−r0 =

∑n≥1,odd ω−r0(n2/8)(−1)

p−1

2 qn2

8

This should be related to the values of the twisted L-series L(ω−r0ψk, k),but we are not aware of the existence of a formula for a “square root” ofthese values.

6 Final p-adic interpolation

We will use the interpolation Iδ(q, s) found in the previous section (definition5.4 and lemma 5.5). We would like to evaluate Iδ(q, s) at an appropriatetest object, to obtain an interpolation for the la’s (defined in 2.2). At thispoint we should pay a closer look at where the values of modular forms lie.We start by quoting some general results.

Let L be a number field, OL its ring of integers, ℘|p in L, L℘ the com-pletion of L at p, O℘ its ring of integers, L℘ the completion of the maximalunramified extension of L℘, O℘ its ring of integers, LN the ray class field(mod N) of L, ℘N |℘ in LN .

28

Theorem 6.1 (Shimura) Let ξ ∈ H, ξ ∈ L an imaginary quadratic field.Let F ∈ Mk(Γ(N), Z[µN ]), G ∈ Mk+2e(Γ(N), Z[µN ]), G(ξ) 6= 0, e, k ∈ Z,e ≥ 0. Then

1. If e = 0 then FG(ξ) ∈ LN .

([34, theorem 3])

2. If FG has integer Fourier coefficients at all cusps, e = 0, and G 6= 0 on

H, then FG(ξ) ∈ OLN

.([34, theorem 3])

3. W eFG (ξ) ∈ LN .

([29, main theorem I] with (1.18) replaced by 1. above)

A similar result holds for half-integral weight.

Lemma 6.2 Let ξ ∈ H, ξ ∈ L an imaginary quadratic field.Let F ∈ Mk(Γ(N), v, Z[µN ]), G ∈ Mk+2e(Γ(N), v′, Z[µN ]), with v2, v′2, vv′

trivial on Γ(N), G(ξ) 6= 0, e ∈ Z, e ≥ 0, but now k ∈ 12Z. Assume there is

θ ∈ M 1

2Z(Γ(N), v, Z[µN ]) with F = F

θ , G = Gθ ∈ MZ(Γ(N), Q[µN ]).

1. If e = 0 then FG(ξ) ∈ LN .

2. If FG has integer Fourier coefficients at all cusps, e = 0, and G 6= 0 on

H, then FG(ξ) ∈ OLN

.

3. W eFG (ξ) ∈ LN .

Proof:1. and 2. F

G(ξ) = FGG2 (ξ) and the theorem applies to FG, G2 ∈

MZ(Γ(N), Z[µN ]).

3. W eFθ = (W + Wθ

θ )eF . Now, Wθθ (ξ) = 1

2Wθ2

θ2 (ξ) ∈ LN by the integral

weight case. Also, W Wθθ = Wθ

θ W + W(

Wθθ

).

Hence, W eFG (ξ) =

(W+Wθθ )

eF

G(ξ) is a sum of terms, each a product of quo-

tients of the required form to apply the previous theorem, part(3).

We keep the notation of section 2. Let p ≥ 5 be a prime that splitscompletely in K = Q(

√−l). Let δ = 0, 1 with l+1

4 ≡ δ (mod 2). Assumethroughout the section that ℘|p is principal. Put ℘ = (µ), ℘ = (µ), µµ = p.Let T be the field obtained by adjoining the values of ψ1 to K. Let ℘T be a

29

prime of T above p. Notice that T is a finite extension, since ψhK1 has values

in K (ψ1 takes values in K on principal ideals).

Lemma 6.3θ2r+ 1

2+δ

η4r+2δ+1(za2) ∈ K24

Proof:θ2r+1

2+δ

η4r+2δ+1 (za2) =(8W )rθ 1

2+δ

η4r+2δ+1 (za2) by lemma 3.12. Apply the previous

Corollary with v = vδ, v′ = vη, N = 24.

Recall the expressions for la(k) and Ωa given in definition 2.2.

Lemma 6.4 The number la(k) lies in a fixed number field.

Proof: This is a corollary to lemma 6.3, since la(2r+δ+1) equalsθ2r+1

2+δ

η4r+2δ+1 (za2)

times an algebraic number involving√

2, (−l)1/4, ψ2r+δ+1(a) and roots ofunity.

Lemma 6.5 (2πi

Ωa

)2r θ2r+δ+ 1

2

θ 1

2+δ

(za2) ∈ K24

Proof:(

2πiΩa

)2r θ2r+δ+1

2

θ 12+δ

(za2) = ζ−2ra

√−l

−r θ2r+δ+1

2

η4r+2δ+1 (za2)η2δ+1

θ 12+δ

(za2) and use

lemma 6.3.

We keep the notation of section 2 and subsection 3.4.

Definition 6.6 A level N complex test object (E, ω, α), (E, ω) ↔ L, ω =Ωdz is said to be defined over a ring A ⊂ C, 1

6 ∈ A, if g2(L), g3(L),℘(Ωα(x, y)) and ℘′(Ωα(x, y)) are in A. (See the isomorphism described inequation (11)).Let (E, ω, α) be defined over A. Let ℘|p be a prime ideal in A. We say that(E, ω, α) has good reduction at ℘ if ∆(E, ω) and ℘ are coprime. We saythat (E, ω, α) has ordinary reduction at ℘ if p(E ⊗ A/℘) ' Z/pZ (the otherpossibility is 0).

30

Let (N, p) = 1. Let La = Ωaza2Z + ΩaZ = Ωa

a2 a2 (see definition 2.2 forΩa). Let (Ea, ωa, αa,N ) be the level N test object

Ea : Y 2 = 4X3 − g2(La)X − g3(La)

ωa = dXY

αa,N (x, y) =(℘(x

Ωaza2

N + yΩa

N ), ℘′(xΩaz

a2

N + yΩa

N )) (15)

where here, and from now on, we omit the (mod La) in the level N struc-ture.

Lemma 6.7 (Gross) The level N test object (Ea, ωa, αa,N ) is defined overOKN

[ 16N1/4 ], has discriminant ∆(Ea, ωa) = −l3 (hence it has good reduction

above p), and has ordinary reduction above p.

Proof: Gross proved that (Ea, ωa) is defined over K1, with ∆(Ea, ωa) = −l3,and Ωa

ΩO∈ K1 (see [7, Ch.5] and [8]). Since Ωa ∈ µ122πi(−l)−1/4η2(za2),

gj(La) ∈ µ3(−l)j/2 gj(za2 )

η2j(za2 )

. This value is integral by virtue of theorem 6.1(2).

So (Ea, ωa) is defined over OK1.

Also, ℘(La

N ), ℘′(La

N ) ∈ OKN[ 1N1/4 ] (see [18, Ch.10.1, thm 2] and [2] for inte-

grality).It is known that if E is a CM-curve with good reduction at p, (E, ω, α) hasordinary reduction if and only if p splits in the field of complex multiplica-tions of E.

Lemma 6.8 Let F ∈ MpZ(Γ(N), Zp[µN ]). Then F(Ea, ωa, αa,N ) ∈ O℘N .

Proof: Since (6N, p) = 1, the previous lemma shows that the test objectis p-integral, hence defined over O℘N . The result now follows immediatelyfrom theorem 4.5.

Lemma 6.9 1.θ2r+δ+1

2

η4r+2δ+1 (za2) ∈ O℘24

2. la(k) is p-integer in a fixed number field.

3.(

2πiΩa

)2r θ2r+δ+1

2

θ 12+δ

(za2) ∈ O℘24

Proof: combine lemmas 6.3, 6.4, 6.5 with lemma 6.8 for integrality.

We see that what remains to be done is:

31

• Evaluate the measure µ∗δ on Zp with values on KMp(Γ(2 − δ), Zp), at

(Ea, ϕa, αa,8) , c−1a as in theorem 4.12, that is: ca ∈ O∗

℘1, c−1

a ωa =

ϕ∗a(

dT1+T ), and αa,8) meaning now the arithmetic level N structure

αa,8)(e2πix/8, y) =′′ old αa,8)

′′(x, y).

• Adjust some constants and prove that the resulting Gamma trans-form is the p-adic function that interpolates the values la(2r + 1 + δ),removing the appropriate Euler factors.

• Put these p-adic functions together to interpolate the values√

L(ψk, k).

Lemma 6.10 The evaluation µ∗δ(Ea, ϕa, αa,8) is a measure on Zp with val-

ues on O℘24, with r-moments, r ≡ 0 (mod p − 1):

(2πica

Ωa

)2r (1 − ε(µ)p2r+δµ−(4r+2δ+1)

) θ 1

2+δ+2r(za2)

θ 1

2+δ(za2)

Proof: For r ≡ 0 (mod p − 1), the moments are:

∫

Zp

xrd (µ∗δ(Ea, ϕa, αa,8)) = Iδ(q, r)(Ea, ϕa, αa,8)

=(8q d

dq )rθ 1

2+δ

θ 1

2+δ

(Ea, ϕa, αa,8) − [(−1)p−1

2 p]δp2r[(8q d

dq )rθ 1

2+δ](q

p2

)

θ 1

2+δ(q)

(Ea, ϕa, αa,8)

by lemma 5.5

=(1 − [(−1)

p−1

2 p]δp2r(−1)p−1

2δε(µ)µ−(4r+2δ+1)

) (8q ddq )rθ 1

2+δ

θ 1

2+δ

(Ea, ϕa, αa,8)

by theorem 7.5 in the Appendix

=(1 − ε(µ)p2r+δµ−(4r+2δ+1)

) (8W )rθ 1

2+δ

θ 1

2+δ

(Ea,Ωa

ca

dz, αa,8)

by equation (14)

=

(ca

Ωa

)2r (1 − ε(µ)p2r+δµ−(4r+2δ+1)

) θ1/2+δ+2r

θ 1

2+δ

(Ea, dz, αa,8)

by lemma 3.12

=

(ca

Ωa

)2r (1 − ε(µ)p2r+δµ−(4r+2δ+1)

) θ 1

2+δ+2r

θ 1

2+δ

(Ea, dz, αa,8)

32

=

(2πica

Ωa

)2r (1 − ε(µ)p2r+δµ−(4r+2δ+1)

) θ 1

2+δ+2r(za2)

θ 1

2+δ(za2)

These moments are in O℘24since ca ∈ O℘1

(theorem 4.12), the Euler Factor

is in OK , and(

2πiΩa

)2r θ 12+δ+2r

(za2 )

θ 12+δ

(za2 ) ∈ O℘24

(see lemma 6.9).

So the measure has values in K℘24, but µ∗

δ has values in KMp(Γ(2− δ), Zp),and (Ea, ϕa, αa,8) is an integral test object, hence µ∗

δ(Ea, ϕa, αa,8) has valuesin O℘24

.Note: The Gamma-transform of µ∗

δ(Ea, ϕa, αa,8) is the functionIδ(q, s)(Ea, ϕa, αa,8).

Next we adjust Iδ(q, s)(Ea, ϕa, αa,8) and µ∗δ(Ea, ϕa, αa,8) to obtain a “par-

tial” interpolation for an ideal class [a].

Lemma 6.11 The expression(

2πiΩa

)δ+ 1

2 θ 1

2+δ(za2) is in OK24

[12 , (−l)1/8] ⊂O℘24

.

Proof: We apply lemma 6.2(1) for 2θ 1

2+δ ∈ Mδ+ 1

2

(Γ(8), vδ, Z[µ8]),

and η2δ+1 ∈ Mδ+ 1

2

(Γ(24), vη, Z[µ24]). Details are left to the reader.

Definition 6.12 Let Hδ,a : Zp −→ O℘24be such that

Hδ,a(s) =

(−4

Na

)δ((

2πi

Ωa

)δ+ 1

2

θ 1

2+δ(za2)

)Iδ(q, s)(Ea, ϕa, αa,8)

We also let

νδ,a =

(−4

Na

)δ((

2πi

Ωa

)δ+ 1

2

θ 1

2+δ(za2)

)µ∗

δ(Ea, ϕa, αa,8)

To simplify the notation, put (EF ) for the Euler Factor:

(EF ) =(1 − ε(µ)p2r+δµ−(4r+2δ+1)

)

Theorem 6.13 With notation as above, we have:

1. Hδ,a(s) =∫

Zp< x >s dνδ,a

33

2. for r ≡ 0 (mod p − 1),

1

c2ra

Hδ,a(r) = (EF )

(2πi

Ωa

)2r+ 1

2+δ (−4

Na

)δ

θ2r+ 1

2+δ(za2)

= (EF )ψ2r+1+δ(a)la(2r + 1 + δ)

Proof: Follows immediatly from lemma 6.10 and its following note.

We have to put together these interpolations to get an interpolation for√L(ψk, k).

Lemma 6.14 On the complex side, we can write

Ωk− 1

2a

ψk(a)= u

k− 1

2a Ω

k− 1

2

OK

for some unit ua ∈ K1.T such that uhKa ∈ O∗

K1→ Z∗

p (this is because ℘ is

principal in K), and u1

2a ∈ O∗

℘1O∗

℘T.

Proof: Put

ua = Ωa

ΩOψ1(a)2

u1

2a =

(Ωa

2πi

) 1

2(

2πiΩO

) 1

2 1ψ1(a)

Lemma 6.7 shows that Ωa

ΩO∈ K1, so ua ∈ K1.T . Also, ψ1(a)

hK ∈ K, since

ahK is principal, and the character takes values in K on principal ideals.Hence uhK

a ∈ K1. Let us prove that ua is a unit. From [18, Corollary p.166],

we get that a24 |∆(a2)|2|∆(OK)|2 is an algebraic unit. Then x =

η2(za2 )

aη2(zO)is an algebraic

unit. For some y ∈ OK , ah = (y).

N(uha) = N

(± xhah

ψ(ah)2

)=

(Nx)ha2h

(Ny)2= (Nx)h

is a unit.Hence ua ∈ O∗

K1.T . Since (a, p) = 1, ψ1(a) is a p-unit. It follows that Ωa/ΩO

is a p-unit, and Ω1

2a /Ω

1

2

O ∈ O∗℘1

(it can only ramify above 2). We deduce that

u1

2a ∈ O∗

℘1O∗

℘T.

34

Lemma 6.15 On the p-adic side, we can write

ca = wacOK

for some unit wa ∈ Z∗p.

Proof: All “p-adic periods” ca satisfy cσa/ca= p-adic unit root of Ea⊗O℘1

/℘1

for σ the Frobenius automorphism of K℘1/K℘1

= Qp/Qp (℘ is principalin K). Since Ea is isogenous to EOK

, their p-adic unit roots are equal.Therefore, cσ

a/ca = cσOK

/cOK. It follows that ca/cOK

∈ Z∗p.

Definition 6.16 For every prime p not dividing h = hK , let Jδ : Zp −→O℘24

O∗℘T

be the function

Jδ(s) =∑

[a]

< uha >2s/h

< wa >2su

δ+ 1

2a Hδ,a(s)

We also let λδ be the measure on Zp with values on O℘24O∗

℘T

νδ =∑

[a]

u1

2+δ

a

((2πi

Ωa

) 1

2+δ

θ 1

2+δ(za2)

) (−4

Na

)δ

µ∗δ

(Ea,

< wa >

< uha >h−1

ϕa, αa,8

)

Good definition: The values are really in O℘24O∗

℘T, according to lemma

6.10, definition 6.12, and lemmas 6.14 6.15.

Theorem 6.17 With notation as above,

1. Jδ(s) =∫

Zp< x >s dνδ

2. for r ≡ 0 (mod h(p − 1)),

1

c2rO

Jδ(r) = (EF )

(2πi

ΩO

)2r+ 1

2+δ √

L(ψ2r+1+δ, 2r + 1 + δ)

eπi8

[22r+δ−2(2r + δ)!π−(2r+1+δ)l−

4r+2δ−1

4

] 1

2

Proof:

35

1.∫

Zp< x >s d

(∑[a] u

1

2+δ

a

((2πiΩa

) 1

2+δ

θ 1

2+δ(za2)

) (−4Na

)δµ∗

δ

)is a form in

KMp2s(Γ0(2 − δ), Zp) (see definition 5.4). Therefore, since <wa>

<uha>h−1 ∈ Z∗

p,

∫

Zp

< x >s dνδ =∑

a

u1

2+δ

a

(< uh

a >h−1

< wa >

)2s ∫

Zp

< x >s dνδ,a

= Jδ(s)

by theorem 6.13(1).2. for r ≡ 0 (mod h(p − 1)):

Jδ(r)

c2rO

=∑

[a]

u2r+ 1

2+δ

a

c2rO w2r

a

Hδ,a(r)

=∑

[a]

u2r+ 1

2+δ

a

c2rO w2r

a

c2ra (EF )

(2πi

Ωa

)2r+δ+ 1

2(−4

Na

)δ

θ2r+δ+ 1

2

(za2)

by theorem 6.13

= (EF )∑

[a]

(−4

Na

)δ (2πi

ΩO

)2r+ 1

2+δ

ψ−(4r+1+δ)1 (a)

(ca

cOwa

)2r

θ2r+ 1

2+δ(za2)

by lemma 6.14

=

(2πi

ΩO

)2r+ 1

2+δ

(EF )∑

[a]

(−4

Na

)δ

ψ−12r+1+δ(a)θ2r+ 1

2+δ(za2)

by lemma 6.15

= (EF )

(2πi

ΩO

)2r+ 1

2+δ √

L(ψ2r+1+δ, 2r + 1 + δ)

eπi8

[22r+δ−2(2r + δ)!π−(2r+1+δ)l−

4r+2δ−1

4

] 1

2

by equation (5)

Note: We have not yet checked whether this interpolation is independentof the isomorphism C ' Cp.

36

7 Appendix: Computation of the Euler Factor

This section is of a technical nature. We do the necessary work to producethe Euler Factor in the final interpolation. We keep the notation of sections2 and 3: K = Q(

√−l), O its ring of integers, l ≡ 3 (mod 4) a positive

prime number, δ = 0, 1 according to l+14 ≡ δ (mod 4), p ≥ 5 a prime

splitting in K, pO = ℘℘. Let K℘ and O℘ be the completions at ℘ of K andO.

Let h, g ⊂ OK be odd ideals, not divisible by rational integers, and(g, h) = 1. Assume that g is principal, g = (%), and gg = gOK . Let h = N(h).

Write h = Zh + Z b+√−l

2 , g = Zg + Z b+√−l

2 .This can be done with the same b mod 2gh, and we can still require thatb ≡ 1 (mod 16).Consider the fractional ideal g−1h. On one hand,

g−1h = ghg−1 = Zh + Zg−1 b +√−l

2

and on the other hand,

g−1h = %−1h = Z%−1h + Z%−1 b +√−l

2

So there must be Mh,g =

(c de f

)∈ SL2(Z) such that:

g−1 b+

√−l

2 = c%−1 b+√−l

2 + d%−1h

h = e%−1 b+√−l

2 + f%−1h(16)

If zh = b+√−l

2h , then g−1zh = czh+dezh+f

Notice that % = ezh + f , is well defined up to a sign, since O∗K = ±1.

Let a ⊂ OK be an odd ideal not divisible by rational integers. Recallthat p splits completely in K, pOK = ℘℘, and assume that ℘ = (µ), ℘ = (µ)are principal, and µµ = p. (The sign of µ is to be determined).We make the following extra assumption about a: (a, 2℘) = 1.Take h = a2, h = a2, and g = ℘, g = p, % = µ. We have proved:

Lemma 7.1 If Ma2,℘ ∈ Γ, and F ∈ Mrk(Γ, v), k ∈ 1

2Z,then

F (p−1za2) = v(Ma2,℘)µkF (za2)

37

Now take a2 as before, and g = ℘2, g = p2, % = µ2.

Lemma 7.2 If Ma2,℘2 ∈ Γ, and F ∈ Mrk(Γ, v), k ∈ 1

2Z,then

F (p−2za2) = v(Ma2,℘2)µ2kF (za2)

Definition 7.3 Let F ∈ MpX(Γ0(N), v, Zp).

Define V (F) ∈ MpX(Γ0(Np), vp, Zp), vp

(a bpc d

)= v

(a pbc d

)by:

V (F)(q) = F(qp)

Good definition: Let Fj −→ F , Fj ∈ Mkj (Γ0(N), v).

Then Fj(pz) ∈ Mkj (Γ0(pN), vp) and Fj(pz) −→ F(qp).

Let (Ea, ωa, αa,R) correspond to (C/Ωaza2Z+ΩaZ, dz, 1RZ) as level R test

objects (see equation (15)), and let ca ∈ O∗℘ be such that c−1

a ωa is magic

(see theorem 4.12). Define ϕa the trivialization satisfying c−1a ωa = ϕ∗( dT

1+T ).The definition of V on Mp

X(Γ0(N), Zp), (N, p) = 1, coincides with Katz’sFrob in [12, 5.5.7]. In particular, if ℘ = (µ) and F ∈ Mp

X(Γ0(N), Zp), wehave V (F) ∈ Mp

X(Γ(N), Zp), and

V (F)(Ea, ϕa, αa,N ) = F(Ea, µϕa, µαa,N ) (17)

See [12, 8.3.8, 8.3.23].

Lemma 7.4 We keep the notation of lemmas 7.1 and 7.2. Assume Ma2,℘ ∈Γ0(N) (resp. Ma2,℘2 ∈ Γ0(N)), F ∈ Mr

k(Γ0(N), v), k ∈ Z, F = zG for some

G ∈ MZ(Γ0(N), v) and some z ∈ Z. Let F = z ? G, F ∈ Mpk(Γ0(N), v).

Assume v(Γ(R)) = 1 for some R divisible by N , so F ∈ Mrk(Γ(R)) and

F ∈ Mpk(Γ(R)).

Then,

V (F)(Ea, ϕa, αa,R) = v(Ma2,℘)µ−kF(Ea, ϕa, αa,R)

V 2(F)(Ea, ϕa, αa,R) = v(Ma2,℘2)µ−2kF(Ea, ϕa, αa,R)

Proof: We prove the first case; the second one is similar.

V (F)(Ea, ϕa, αa,R) = F(Ea, µϕa, µαa,R)

by equation (17), since F ∈ MpZ(Γ(R))

38

= F(Ea, µc−1a ωa, µαa,R)

by definition 4.11

= (µc−1a )−kF(Ea, ωa, µαa,R)

by theorem 4.5

Then by theorem 4.14,

V (F)(Ea, ϕa, αa,R) = (Ωac−1a µ)−kF (C/za2Z + Z, dz,

µ

RZ)

We can keep track of the changes introduced by the use of the new level Rstructure. We saw that

℘−1a2 = Za2 + Zb +

√−l

2p

Therefore,

a2 = µ℘−1a2 = Zµa2 + Zµ−1 b +√−l

2So,

V (F)(Ea, ϕa, αa,R) = (Ωac−1a µ)−kF (C/za2Z + Z, dz,

µ

RZ)

= (Ωac−1a µ)−kF (C/Zµ + Zµ−1za2 , dz,

µ

RZ)

= (Ωac−1a µ)−kF (µ−1za2 , µ)

by definition 3.17

= (Ωac−1a µ)−k

(2πi

µ

)k

F (p−1za2)

by equation (10)

We now use lemma 7.1:

V (F)(Ea, ϕa, αa,R) = (Ωac−1a µ)−k

(2πi

µ

)k

v(Ma2,℘)µkF (za2)

= v(Ma2,℘)µkF(Ea, ϕa, αa,R)

Theorem 7.5 With the same notation as above with R = 8,[(8q d

dq

)rθ 1

2+δ

](qp2

)

θ 12+δ

∈ KMp2r(Γ(2 − δ), Zp) and

[(8q d

dq

)rθ 1

2+δ

](qp2

)

θ 1

2+δ

(Ea, ϕa, αa,8) =

39

(−1)p−1

2δε(µ)µ−(4r+1+2δ)

(8q d

dq

)rθ 1

2+δ

θ 1

2+δ

(Ea, ϕa, αa,8)

Proof:

Lemma 5.3 shows that

[(8q d

dq

)rθ 1

2+δ

](qp2

)

θ 12+δ

∈ KMp2r(Γ(2 − δ), Zp).

[(8q d

dq

)rθ 1

2+δ

](qp2

)

θ 1

2+δ

(Ea, ϕa, αa,8) =

= V 2

(8q d

dq

)rθ 1

2+δ

θ 1

2+δ

(Ea, ϕa, αa,8)

θ 1

2+δ(q

p2

)

θ 1

2+δ(q)

(Ea, ϕa, αa,8)

= µ−4r

(8q d

dq

)rθ 1

2+δ

θ 1

2+δ

(Ea, ϕa, αa,8)θ 1

2+δ(q

p2

)

θ 1

2+δ(q)

(Ea, ϕa, αa,8)

by virtue of lemma 7.4, applied toθ2r+1

2+δ

θ 12+δ

∈ Mp2r(Γ0(2 − δ)) (v = 1 here).

(We check that Ma2,℘2 =

(c de f

)∈ Γ0(2− δ): we saw that µ2 = eza2 + f ,

so Im(µ2) = e2a

√l ∈ Z

√l, and we get that e is even.)

It remains to evaluate the second factor.

θ 1

2+δ(q

p)

θ 1

2+δ(q)

(Ea, ϕa, αa,8)

2

=θ2

1

2+δ

(qp)(Ea, ϕa, αa,8)

θ21

2+δ

(q)(Ea, ϕa, αa,8)

= v2δ (Ma2,℘)µ−(1+2δ),

since θ21

2+δ

∈ Mp1+2δ(Γ0(2 − δ), v2

δ ), v2δ (Γ0(4)) = 1, and lemma 7.4 applies.

(We check that Ma2,℘ =

(c de f

)∈ Γ0(2 − δ): we saw that µ = eza2 + f ,

so Im(µ) = e2a

√l ∈ Z

√l, and we get that e is even.)

We deduce:

θ 1

2+δ(q

p)

θ 1

2+δ(q)

(Ea, ϕa, αa,8) = ±vδ(Ma2,℘)µ−( 1

2+δ) (18)

40

A short calculation shows that

θ 1

2+δ(pz)

θ 1

2+δ(z)

∈ M0(Γ0((2 − δ)p), wp) ⊂ M0(Γ(8p))

where wp(γ) = vδ(γp)/vδ(γ), and and γp =

(a pbc d

)if γ =

(a bpc d

).

In any case,

θ 1

2+δ(q

p)

θ 1

2+δ(q)

=

˜

θ 1

2+δ(pz)

θ 1

2+δ(z)

∈ KMp

0(Γ(8p), Zp) ⊂ KMp(Γ(8), Zp)

(see theorem 4.8).We apply Shimura reciprocity. In this context, it says that the evaluation of

V (F) equals σ of the evaluation of F , where σ is an extension of(

℘K(8℘∞)/K

)

(see [12, lemma 8.3.25]). We get:

θ 1

2+δ(q

p2

)

θ 1

2+δ(q)

(Ea, ϕa, αa,8) = (V + 1)

θ 1

2+δ(q

p)

θ 1

2+δ(q)

(Ea, ϕa, αa,8)

= (σ + 1)

θ 1

2+δ(q

p)

θ 1

2+δ(q)

(Ea, ϕa, αa,8)

= (σ + 1)(±vδ(Ma2,℘)µ−( 1

2+δ))

according to equation (18)

= vδ(Ma2,℘)p+1σ(µ1

2 )µ−(1+2δ)

Claim: by adjusting b, we can choose f > 0 and f ≡ 1 (mod 4), and, inthat case, vp+1

δ (Ma2,℘) = 1.(Note: recall that b is defined mod 2pa2. If we change b by b + 2pa2r forsome r ∈ Z, f gets replaced by f + pr.)

Proof: going back to (16), we have Ma2,℘ =

(c de f

)∈ SL2(Z) such that:

p−1 b+

√−l

2 = cµ−1 b+√−l

2 + dµ−1a2

a2 = eµ−1 b+√−l

2 + fµ−1a2(19)

41

The second equation says µ = eza2 + f . Choose f > 0. That fixes µ.Claim: we can choose µ, or, equivalently, f , such that,

Ma2,℘ ≡

(1 00 1

)(mod 4) if p ≡ 1 (mod 4)

(1 02 1

)(mod 4) if p ≡ 3 (mod 4) and δ = 0

(1 22 1

)(mod 4) if p ≡ 3 (mod 4) and δ = 1

In all cases, vp+1δ (Ma) = 1 (notice that Ma2,℘ ∈ Γ0(2) and f > 0, so we can

use the formulas for vδ in (8)and (9)).Proof: From (19):

Re(µ) = eb2a2 + f

Im(µ) =√

l e2a2

(Re(µ) − cp) b2 − Im(µ)√

ll2 = dpa2

Im(µ)√l

b + Re(µ) − cp = 0

Notice that, since µ ∈ Z + Z1+√−l

2 , Im(µ)√l

∈ Z. Also√−l = i

√l.

Then, since a2 ≡ 1 (mod 8),

e2 ≡ Im(µ)√

l(mod 8)

f ≡ Re(µ) − Im(µ)√l

(mod 8)

cp ≡ Im(µ)√l

+ Re(µ) (mod 8)

dp ≡ − Im(µ)√l

b2+l2 (mod 8)

(20)

Our choice of b ≡ 1 + 2pr (mod 16), 2a2 ≡ 2 (mod 16), and p odd,combined with the previous equations, imply that, first e is even and e/2(mod 8) is independent of a, then f (mod 8) is independent of a,c (mod 8) is independent of a, and finally d (mod 8) is independent of a.Take norm of µ. We get

p ≡ e2 b2 + l

4+ ef + f2 (mod 8)

Since e is even and p is odd, f has to be odd. Recall that b2+l4 ∈ Z, and

hence p ≡ ef + 1 (mod 4).

42

Therefore, e ≡ 0 (mod 4) if p ≡ 1 (mod 4), and e ≡ 2 (mod 4) if p ≡ 3(mod 4).We have Re(µ) and Im(µ)/

√l determined by µ.

At this point, let us choose r. We saw that f corresponding to the original bis odd. Therefore, f ≡ ±1 (mod 4). If f ≡ 1 (mod 4), we do not changeit, that is, we take r = 0. If f ≡ −1 (mod 4), f + pr ≡ −1 + pr (mod 4).We want it to be 1 (mod 4). We solve for r, and we immediately see thatr = 2t, t is odd, and the new b can be taken congruent to 1+4p (mod 16).We make sure to take r > 0, in order not to ruin the condition f > 0.In summary, from now on, f > 0, f ≡ 1 (mod 4), b ≡ 1, 1 + 4p (mod 8).

Notice, in both cases, b2 ≡ 1 (mod 8) and b2+l4 ≡ 1+l

4 (mod 2).

From (20), cp ≡ 2 Im(µ)√l

+ f (mod 8), hence cp ≡ 2 e2 + f (mod 8). So:

cp ≡ e + f (mod 8)

We deduce that c ≡ f ≡ 1 (mod 4) for all p. Again from (20),

dp ≡ −el + b2

4(mod 8)

If δ = 0, l+14 and b2+l

4 are even. Hence d ≡ 0 (mod 4). If δ = 1, l+14 and

b2+l4 are odd. We obtain d ≡ 0 (mod 4) if p ≡ 1 (mod 4), and d ≡ 2

(mod 4) if p ≡ −1 (mod 4).

Claim: σ(µ1

2 ) = (−1)p−1

2δε(µ).

Proof:•σ(µ

1

2 ) =(

µ+µp

).

σ acts as(

℘K(8℘∞)/K

)on µ

1

2 . So we need to know whether µ is a square

(mod µ). But that is equivalent to deciding whether µ + µ is a square (modµ). Since µ + µ ∈ Z and p splits in K, that is measured by the quadratic

symbol(

µ+µp

). Notice that this symbol is symmetric on µ and µ.

• by definition, ε(µ) =(

2(µ+µ)l

).

•µ + µ = eba2 + 2f , in the previous notation.

•µ + µ = 2r(−1)hk, with k > 0 and odd. We need some informationabout r.Claim:

If p ≡ 1 (mod 4), r = 1.

43

If p ≡ 3 (mod 4), r = 2, except when v2(e − 2) = 2, and in that case,(2l

)=

(2p

).

Proof:Assume p ≡ 1 (mod 4). Then e ≡ 0 (mod 4). Recall f ≡ 1 (mod 4).Hence,eba2 + 2f ≡ e + 2 ≡ 2 (mod 4). So r = 1.Now assume p ≡ 3 (mod 4). Recall b ≡ 1, 1 + 4p (mod 16), e ≡ 2

(mod 4), f ≡ 1 (mod 4). Also note that(

2l

)= (−1)δ.

Put e = 2 + 2st, s ≥ 1, t odd. Then, eba2 + 2f ≡ 4 + 2s5t, 4 + 2st (mod 8).

If s 6= 2, we deduce that r = 2. If s = 2, e = 2 + 4t, t odd. Hence e ≡ 6(mod 8). Replace in the norm formula:

p = e2 b2 + l

4a2+ f2 +

ebf

a2, (21)

obtaining: p ≡ 4 b2+l4 + 1 + 6 ≡ (1 + l) + 7 ≡ 4δ + 7 (mod 8). Then,(

2l

)= (−1)δ = (−1)

p−7

4 =(

2p

).

We have,

ε(µ) =

(2(µ + µ)

l

)=

(2

l

)r+1(−1

l

)h (k

l

)=

(2

l

)r+1

(−1)h(−1)l−1

2

k−1

2

(l

k

)

Now, from (21) and µ + µ = eba2 + 2f , we obtain

(µ + µ)2 = 4p − e2l

4.

Then,

ε(µ) =

(2

l

)r+1

(−1)h+ k−1

2

(4p − (µ + µ)2

k

)

=

(2

l

)r+1

(−1)h+ k−1

2

(p

k

)

=

(2

l

)r+1

(−1)h+ k−1

2 (−1)p−1

2

k−1

2

(k

p

)

=

(2

l

) ((2

l

) (2

p

))r

(−1)h+ k−1

2(1+ p−1

2)(−1

p

)h (µ + µ

p

)

If p ≡ 3 (mod 4), we get ε(µ) = (−1)δσ(µ).

44

If p ≡ 1 (mod 4),

ε(µ) = (−1)h+ k−1

2

(2

p

)σ(µ).

In this case, p ≡ e + 1 (mod 8), k = (−1)h(

eb2a2 + f

), e ≡ 0 (mod 4).

Then,

k − 1 ≡ (−1)h e

2− 2h (mod 4),

and it follows that h + k−12 ≡ e

4 (mod 2). Since p2−18 ≡ e

4 (mod 2),

(−1)h+ k−1

2

(2p

)= 1, and we get ε(µ) = σ(µ).

In all cases,

ε(µ) = (−1)δ p−1

2 σ(µ).

This ends the proof of the theorem.

Note: ε(µ)µ is equal to the p-adic unit root of EO ⊗O℘1/℘1.

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48