# Using Continued Fractions to Compute Iwasawa Lambda Invariants of Imaginary Quadratic Number Fields

Authors: Jordan Schettler
Journal: Journal of Numbers
 (2014)

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## Abstract

Letl>3 be a prime such that l≡3(mod4) and Q(l) has class number 1. Then Hirzebruch and Zagier noticed that the class number of Q(-l) can be expressed as h(-l)=(1/3)(b1+b2+⋯+bm)-m where the bi are partial quotients in the “minus” continued fraction expansion l=[[b0;b1,b2,…,bm¯]]. For an odd prime p≠l, we prove an analogous formula using these bi which computes the sum of Iwasawa lambda invariants λp(-l)+λp(-4) of Q(-l) and Q(-1). In the case that p is inert in Q(-l), the formula pleasantly simplifies under some additional technical assumptions.


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## Body

## 1. Notation and Assumptions

LetK be a real quadratic number field of discriminant D. Suppose
(1)D=D1D2,
where D1, D2 are the discriminants of quadratic number fields K1, K2, respectively. We will frequently make the following assumption.Assumption A.
Suppose the class number ofK is 1 and that D is divisible by a prime congruent to 3 modulo 4.Remark 1.
Make Assumption A. ThenK has no units of negative norm and the factorization in (1) is unique (up to ordering of factors) with D1, D2 negative by classical genus theory. Without loss of generality, -D1 is a prime congruent to 3 modulo 4 and -D2 is either 4, 8, or a prime congruent to 3 modulo 4.Fori=1,2 let h(Di) denote the class number of Ki. For a prime p, let λp(Di) denote the Iwasawa lambda invariant of the cyclotomic Zp-extension of Ki.Goal. Under Assumption A, we want a formula for the sum of lambda invariants λp(D1)+λp(D2) which is analogous to Hirzebruch and Zagier’s formula for the product of class numbers h(D1)h(D2) given in terms of the partial quotients in the “minus” continued fraction expansion of (δ+D)/2 where δ∈{0,1} with D≡δ(mod⁡4).To accomplish this goal, we first recall some computations of special values of partial zeta functions obtained by Kronecker limit formulas ats=1 or by the methods of Takuro Shintani at s=0. Then we relate these to special values of L-functions which can be alternatively given in terms of the arithmetic invariants h(Di) and λp(Di).


## 2. Special Values of Partial Zeta Functions

Supposem=m0m∞ is a modulus of K where we view m0 as an ideal in the ring of integers OK and we will always assume m∞ is the product of both real places of K. We denote the narrow ray class group associated with m by Cm as in [1]. Consider the partial zeta function ζ(s,C) associated with some C∈Cm, that is, the meromorphic continuation of the sum
(2)∑a∈Ca⊆OK1Nas,
where N(a) denotes the absolute norm of a. We have a Laurent expansion
(3)ζs,C=κs-1+ϱC+ϱ1Cs-1+ϱ2Cs-12+⋯,
where κ is a constant which depends on m but not on C. Computations of ϱ(C) are called “Kronecker limit formulas” for real quadratic number fields because Leopold Kronecker first computed this quantity in the context of an imaginary quadratic number field. If χ is a nontrivial character on Cm, the L-function
(4)Ls,χ=∑C∈CmχCζs,C
has special values at s=1,0 given by
(5)L1,χ=∑C∈CmχCϱC,L0,χ=∑C∈CmχCζ0,C.
We will state some results which express ϱ(C), ζ(0,C) in terms of continued fractions, and, in order to do so, we need a couple of lemmas which we do not prove here. See [2, 3].Lemma 2.
Supposeα∈R∖Q. There is a unique “minus” continued fraction expansion (this “minus” expansion is related to the usual “plus” continued fraction expansion α=[a0;a1,a2,a3,…] where a0∈Z and a1,a2,…∈Z>0: as sequences (b0,b1,b2,…) = (a0+1,2,2,…,2︸a1-1,a2+2,2,2,…,2︸a3-1,a4+2,…))(6)α=b0;b1,b2,b3,…≔lim⁡k→∞b0-1b1-1/b2-⋯-1/bk,
where b0∈Z and b1,b2,…∈Z>1. Moreover, this expansion is eventually periodic if and only if α is algebraic of degree 2. In particular, for δ∈{0,1} with D≡δmod⁡4,
(7)δ+D2=b0;b1,b2,…,bm¯,
where the bar signifies the repeating block of minimal length m; moreover, bm=2b0-δ and we have a palindrome
(8)b1,b2,…,bm-1=bm-1,bm-2,…,b1.Lemma 3.
LetC∈Cm. For each integral ideal a∈C, there are totally positive z,ϖ∈K such that
(9)za-1m0=Z+ϖZ,
where ϖ>1>ϖ′>0 with ϖ′ = Galois conjugate of ϖ. This condition on ϖ ensures that its “minus” continued fraction is purely periodic of some minimal period m:
(10)ϖ=b0;b1,b2,…,bm-1¯.
Moreover, the sequence (b0,b1,…,bm-1) is determined up to cyclic permutation by C.


### 2.1. Meyer’s Theorem

Meyer studied the casem0=OK in [4]. He expressed ϱ(C) as an integral involving a logarithm of the Dedekind η-function using methods earlier applied by Erich Hecke to wide ideal classes. The advantage of considering narrow ideal classes is that if there are no units of negative norm, every wide ideal class is the disjoint union of two narrow ideal classes C∐‍C*, and the transformation properties of the Dedekind-η function can then be used to explicitly evaluate the difference ϱ(C)-ϱ(C*) as π2/D multiplied times an expression involving a Dedekind sum. Friedrich Hirzebruch and Don Zagier noticed that this expression could be written as a sum of partial quotients in a certain “minus” continued fraction.Definition 4.
We denote by1+m0 the set of α∈K× such that ordp(α-1)≥ordp(m0) for all prime ideals p of OK dividing m0. Define Θ=[(θ)]∈Cm where θ is any positive element of 1+m0 whose Galois conjugate θ′ is negative. For each C∈Cm, we take C*=CΘ.Theorem 5 (Meyer, as restated by Hirzebruch and Zagier).
Supposem0=OK and that K has no units of negative norm. In the notation of Lemma 3, we have
(11)ϱC-ϱC*=π26D∑k=1mbk-3.


### 2.2. Yamamoto’s Theorem

Meyer’s theorem is sufficient to derive the known formula for class numbers, and there are generalizations which computeϱ(C), ϱ(C*) for an arbitrary m0. For example, Yamamoto proved such a Kronecker limit formula for narrow ray classes in [5]; he further computed ζ(0,C), ζ(0,C*) using the methods of Shintani in [6]. We will find it more convenient to use these computations at s=0 because they are rational numbers and have considerably simpler descriptions in the general case.Definition 6.
LetC∈Cm. Choose a, z, ϖ=[[b0;b1,…,bm-1¯]] as in Lemma 3. There are unique rational numbers c-2,c-1∈[0,1) such that
(12)c-2-c-1ϖ-z∈Z+ϖZ,
so if we extend bk=bk+m by periodicity, we may recursively define ck∈[0,1) for all integers k≥0 by
(13)ck=bkck-1-ck-2,
where x=x-x denotes the fractional part of a real number x.Definition 7.
LetU+ denote the totally positive units in OK with generator ε>1. Take εm to be the unique generator of U+∩(1+m0) which is greater than 1, so εm=εr for some nonnegative integer r.Theorem 8 (Yamamoto).
LetC∈Cm. Then
(14)ζ0,C=-ζ0,C*=∑k=1rmbk2B2ck-1-B1ck-1B1ck-2,
where B1(x)=x-1/2, B2(x)=x2-x+1/6 are Bernoulli polynomials and bk, m, ck, r are as in Definitions 6, 7.


## 2.1. Meyer’s Theorem

Meyer studied the casem0=OK in [4]. He expressed ϱ(C) as an integral involving a logarithm of the Dedekind η-function using methods earlier applied by Erich Hecke to wide ideal classes. The advantage of considering narrow ideal classes is that if there are no units of negative norm, every wide ideal class is the disjoint union of two narrow ideal classes C∐‍C*, and the transformation properties of the Dedekind-η function can then be used to explicitly evaluate the difference ϱ(C)-ϱ(C*) as π2/D multiplied times an expression involving a Dedekind sum. Friedrich Hirzebruch and Don Zagier noticed that this expression could be written as a sum of partial quotients in a certain “minus” continued fraction.Definition 4.
We denote by1+m0 the set of α∈K× such that ordp(α-1)≥ordp(m0) for all prime ideals p of OK dividing m0. Define Θ=[(θ)]∈Cm where θ is any positive element of 1+m0 whose Galois conjugate θ′ is negative. For each C∈Cm, we take C*=CΘ.Theorem 5 (Meyer, as restated by Hirzebruch and Zagier).
Supposem0=OK and that K has no units of negative norm. In the notation of Lemma 3, we have
(11)ϱC-ϱC*=π26D∑k=1mbk-3.


## 2.2. Yamamoto’s Theorem

Meyer’s theorem is sufficient to derive the known formula for class numbers, and there are generalizations which computeϱ(C), ϱ(C*) for an arbitrary m0. For example, Yamamoto proved such a Kronecker limit formula for narrow ray classes in [5]; he further computed ζ(0,C), ζ(0,C*) using the methods of Shintani in [6]. We will find it more convenient to use these computations at s=0 because they are rational numbers and have considerably simpler descriptions in the general case.Definition 6.
LetC∈Cm. Choose a, z, ϖ=[[b0;b1,…,bm-1¯]] as in Lemma 3. There are unique rational numbers c-2,c-1∈[0,1) such that
(12)c-2-c-1ϖ-z∈Z+ϖZ,
so if we extend bk=bk+m by periodicity, we may recursively define ck∈[0,1) for all integers k≥0 by
(13)ck=bkck-1-ck-2,
where x=x-x denotes the fractional part of a real number x.Definition 7.
LetU+ denote the totally positive units in OK with generator ε>1. Take εm to be the unique generator of U+∩(1+m0) which is greater than 1, so εm=εr for some nonnegative integer r.Theorem 8 (Yamamoto).
LetC∈Cm. Then
(14)ζ0,C=-ζ0,C*=∑k=1rmbk2B2ck-1-B1ck-1B1ck-2,
where B1(x)=x-1/2, B2(x)=x2-x+1/6 are Bernoulli polynomials and bk, m, ck, r are as in Definitions 6, 7.


## 3. The Formula for Class Numbers

In this section, we use the results stated above to outline a proof of the formula for class numbers due first in a special case to Hirzebruch (in [7]) and then in the more general form below to Zagier (in [2]). We do this in order to motivate the formula for Iwasawa lambda invariants.We sayχ is a genus character when χ is a real valued character on Cm with m0=OK. In this case, we can use either Meyer’s Theorem 5 or Yamamoto’s Theorem 8 to compute class numbers by factoring L(s,χ) into the product of Dirichlet L-functions.Theorem 9 (Kronecker).
Lett denote the number of distinct prime factors of D. Then there are exactly 2t-1-1 ways to factor D=D1D2 up to order as in (1). Such factorizations are in bijection with the set of nontrivial genus characters χ. Under this correspondence,
(15)Ls,χ=Ls,ϵ1Ls,ϵ2,
where each ϵi is the quadratic character of Ki/Q=Q(Di)/Q.Theorem 10 (Hirzebruch).
Make Assumption A. Then(16)hD1hD2=w1w224∑k=1mbk-3,
where (δ+D)/2=[[b0;b1,b2,…,bm¯]] as in Lemma 2 and each wi is the number of roots of unity in Ki.Proof.
Takem0=OK. By Remark 1 and Theorem 9, there is a unique nontrivial genus character χ on Cm, so (15) and the analytic class number formula imply
(17)L1,χ=2πhD1w1-D1·2πhD2w2-D2=4π2w1w2DhD1hD2
or via functional equations
(18)L0,χ=2hD1w1·2hD2w2=4w1w2hD1hD2.
Here we have simply Cm={Θ,Θ*}. Consider the trivial class C=Θ* in the context of Lemma 3: we can choose a=OK, z=1, and
(19)ϖ=2b0-δ+D2=2b0-δ,b1,b2,…,bm-1¯
via Lemma 2. It is also clear that c-2=c-1=0 as in Definition 6, so ck=0 for all k. Thus Meyer’s Theorem 5 implies
(20)L1,χ=ϱΘ*-ϱΘ**=π26D∑k=1mbk-3
and Yamamoto’s Theorem 8 implies
(21)L0,χ=ζ0,Θ*-ζ0,Θ**=16∑k=1mbk-3.
Combining either (20) with (17) or (21) with (18) will  yield the desired result.


## 4. The Formula for Iwasawa Lambda Invariants

Fix a primep and number field F. Let F∞ denote the cyclotomic Zp-extension of F (we will not consider any noncyclotomic Zp-extensions in this paper); that is, F∞ is the unique subfield of
(22)⋃n≥1Fζpn⊆Q¯
such that Gal(F∞/F) is isomorphic to the group Zp of p-adic integers where Q¯ is some fixed algebraic closure and each ζpn a primitive pnth root of unity. The subfields of F∞ which contain F all lie in a tower
(23)F⊂F1⊂F2⊂⋯⊂F∞,
where
(24)GalFnF≅Zpn∀n≥1.
The p-parts of the class numbers of these intermediate fields become regularly behaved.Theorem 11 (Iwasawa’s growth formula).
There are integersλp(F), μp(F), νp(F) such that class numbers hn of Fn satisfy
(25)ordphn=λpFn+μpFpn+νpF
for all sufficiently large n where ordp denotes the p-adic order.Here is a short list of what is known and conjectured about the Iwasawa invariantsλ, μ, ν which appear in the growth formula.(i)
Iwasawa conjectured thatμp(F)=0 for all p and F.(ii)
Ferrero and Washington proved thatμp(F)=0 for all p when F/Q is abelian (see [8]).(iii)
IfF has only one prime lying over p and p does not divide the class number of F, then λp(F)=μp(F)=νp(F)=0 (see [9]).(iv)
Ifp splits completely in F, then λp(F)≥r2 where r2 is the number of complex places of F (see, e.g., [10]).(v)
Greenberg conjectured thatλp(F)=0 for all primes p when F is a totally real number field (see [11]).Suppose now that F is a quadratic number field of discriminant Δ, and write
(26)λpΔ≔λpF,μpΔ≔μpF.
Thus we always have μp(Δ)=0, and conjecturally λp(Δ)=0 when Δ>0. Assume now that Δ<0. Then λp(Δ)≥1 for infinitely many primes p. Using the analogy between number fields and function fields over finite fields, Ralph Greenberg (after Theorem  3.3 in [10]) suggests the possibility that for a fixedΔ_ the lambda invariants λp(Δ) could be bounded as p varies over all primes; however, neither the boundedness nor unboundedness of the set {λp(Δ):pprime} is known for any single Δ<0. On the other hand, it is conjectured that λp(Δ) is unbounded for fixedp_ as Δ ranges over all discriminants of imaginary quadratic number fields. Ferrero (see [12]) and Kida (see [13]) proved that for -4≠Δ≠-8 we have
(27)λ2Δ=-1+∑l∣Δl≠22ord2l2-1-3,
where the sum ranges over all odd primes l dividing Δ. In particular, this shows that λ2(Δ) is unbounded. For odd p, there seems to be no simple formula like (27) to compute λp(Δ). We will derive a formula for λp(D1)+λp(D2) under Assumption A which is analogous to the formula (16) for class numbers. We first need to recall how the lambda invariant in the growth formula (25) is related to special values of L-functions. We assume here that p∤Δ and p is odd for simplicity. Let ϵ denote the quadratic character for F/Q=Q(Δ)/Q. For each integer n≥1, choose a Dirichlet character ψn of conductor pn+1 and order pn; equivalently, ψn generates the nth level Qn⊆Q(ζpn+1) in the cyclotomic Zp-extension of Q. By a theorem of Kubota and Leopoldt, there is a p-adic analytic function Lp(s,ϵψnω) on the disk s<p(p-2)/(p-1) in Cp such that
(28)Lp1-m,ϵψnω=1-ϵψnω1-mppm-1L1-m,ϵψnω1-m
for all integers m≥1 where ω is the Teichmüller character. In fact, there is an interpolating power series f(T,ϵω)∈Zp[[T]] such that
(29)Lps,ϵψnω=fζpn1-pΔs-1,ϵω
for all s∈Zp where ζpn=ψn(1-pΔ)-1 is a primitive pnth root of unity. Setting 0=1-m=s in the previous two equations gives
(30)fζpn-1,ϵω=Lp0,ϵψnω=L0,ϵψn.
We define lambda and mu invariants of the power series
(31)fT,ϵω=a0+a1T+a2T2+a3T3+⋯
as follows:
(32)μf≔min⁡⁡ordpai:i≥0,λf≔min⁡⁡i≥0:ordpai=μf.
On can use (30) to prove the growth formula (25) for F=Q(Δ) (see, e.g., [14] or much earlier [15]), and, in fact,
(33)μf=0,λf=λpΔ.
Here we are using the assumption that p is odd; we get a different computation for μ(f) when p=2. We compute
(34)ordpL0,ϵψn=ordpa0+a1ζpn-1+a2ζpn-12+⋯≥min⁡⁡ordpai1-ζpni:i≥0=min⁡⁡ordpai+iφpn:i≥0,
where φ is the Euler totient function. The inequality is an equality if the minimum is assumed by exactly one member of the set. In particular,
(35)ordpL0,ϵψn=λpΔφpn
whenever φ(pn)>λp(Δ). Note that we always have
(36)ordpa0+a1ζpn-1+⋯ζpn-1λpΔ-1m+aλpΔ-1ζpn-1λpΔ-1≥1,ordpaλpΔζpn-1λpΔ+aλpΔ+1m·ζpn-1λpΔ+1+⋯=λpΔφpn.
Thus letting pn=(1-ζpn) denote the unique prime ideal lying above p in Z[ζpn], we get
(37)λpΔ=φpnordpL0,ϵψn=ordpnL0,ϵψn
whenever ordpnL(0,ϵψn)<φ(pn). There is a partial converse to this statement which follows from the same observations; namely, if ordpnL(0,ϵψn)≥φ(pn), then we must also have λp(Δ)≥φ(pn).At this point, we should remark that the special valuesL(0,ϵψn) can be computed with generalized Bernoulli numbers via the formula
(38)L1-m,ϵψn=-Bm,ϵψnm∀integersm≥1.
In particular,
(39)L0,ϵψn=-B1,ϵψn=-1f∑a=1fϵaψnaa,
where f=-Δpn+1 is the conductor of ϵψn. This shows that L(0,ϵψn) is an algebraic integer by the work of Carlitz in [16]. However, we will compute this special value in a different way by using Yamamoto’s Theorem 8.FactorD=D1D2 as in (1), and suppose each Di<0. As above, let ϵi denote the quadratic character for Q(Di)/Q. Then (37) implies that for sufficiently large n (which we fix for the following discussion) we have
(40)λpD1+λpD2=ordpnL0,ϵ1ψn+ordpnL0,ϵ2ψn=ordpnL0,χn=ordpn∑C∈CmχnCζ0,C,
where
(41)Ls,χn=Ls,ϵ1ψnLs,ϵ2ψn
is the L-function for a character χn on the narrow ray class group Cm of the real quadratic number field K=Q(D) with modulus m=(pn+1)m∞. For a prime ideal q of OK with q∩Z=(q) and q≠p we have
(42)χnq=χqψnqf,
where χ is the nontrivial genus character associated with the factorization D=D1D2 and f is the residue degree of q over q. Thus for a nonzero ideal I in OK we have χn(I)=χ(I)ψn(N(I)) where N(I) is the absolute norm of I, so χn((a))=ψn(a2) for all a∈Z. Suppose now that D is divisible by a prime congruent to 3 modulo 4. Then the narrow ray class group Cm is an internal direct product Cm+×Θ where Θ is as in Definition 4 and Cm+ is the kernel of the natural homomorphism Cm→Cm∞≅Z/(2). We have χn(Θ)=-1, so χn(C)ζ(0,C)=χn(C*)ζ(0,C*) for all C∈Cm. Thus since ordpn(2)=0 we get
(43)λpD1+λpD2=ordpn∑C∈Cm+χnCζ0,C.
Suppose, in addition, that the class number of K is 1. We claim there is an exact sequence
(44)U+=ε⟶OKpn+1×⟶Cm+⟶0,
where the first map sends the fundamental unit ε to its congruence class ε¯ modulo pn+1 and the second map sends the congruence class α¯ modulo pn+1 of a totally positive α∈OK to the class [(α)]∈Cm+ of the principal ideal (α)⊆OK. (Note that every congruence class modulo pn+1 has a totally positive representative, and any two such representatives for the same congruence class will generate the same narrow ray class.) It is clear that the second map in (44) onto since every class in Cm+⊆Cm is represented by an integral ideal I prime to m0=(pn+1), but such an ideal must be principal I=(α) (since K has class number 1) with totally positive generator α (since the class of I is trivial in Cm∞). Suppose [(α)] is trivial in Cm+ for some totally positive α∈OK relatively prime to p. Then (α)=(β) for some totally positive β∈1+m0=1+(pn+1), but we can write β=a/b for some a,b∈OK relatively prime to p. Since both α, β are totally positive, we have α=εkβ=εka/b for some k≥0. Hence b¯·α¯=ε¯k·a¯ in (OK/(pn+1))×, but a¯=b¯ since β=a/b∈1+(pn+1), so α¯=ε¯k is in the image of the first map. This proves exactness of the sequence in (44) as claimed.Now consider a fixed class[(α)] for some totally positive α∈OK relatively prime to pn+1. In the context of Lemma 3 with m0=(pn+1), we may take a=(α) and z=α/pn+1 so that
(45)za-1m0=OK=Z+ϖZ
with
(46)ϖ=2b0-δ+D2=2b0-δ,b1,b2,…,bm-1¯,
where (δ+D)/2=[[b0,b1,…,bm¯]] as in Lemma 2. Write
(47)α=x+yδ+D2
with x,y∈Z. Define
(48)c-2=x-b0-δypn+1,c-1=-ypn+1,
so that condition (12) is satisfied. As per (13), we have
(49)c0=2b0-δc-1-c-2=-x+b0ypn+1,
and it follows that for all k≥-1(50)ck=-xqk+ypkpn+1,
where pk and qk are the numerator and denominator, respectively, of the kth convergent [[b0;b1,…,bk]] for (δ+D)/2 with p-1=1, q-1=0 by convention. Then Yamamoto’s Theorem 8 implies
(51)ζ0,α=∑k=1rnmbk2B1ck-12-B1ck-1B1ck-2+C,
where rn=log⁡⁡εm/log⁡⁡(ε) is the order of ε modulo pn+1 and
(52)C=-rn24∑k=1mbk
does not depend on the class [(α)].Letg∈Z be a primitive root modulo all powers of p, so, in particular, g¯ has order pn(p-1) in (OK/(pn+1))×. Let r0 denote the order of ε modulo p. Then r0∣p±1 where the sign is + or − when p is inert or split, respectively, in K. We have an isomorphism of abelian groups
(53)OKpn+1×≅Zpnp2-1⊕Zpnpinert inKZpnp-1⊕Zpnp-1psplit inK.
Note that if εa≡b(mod⁡pn+1) for some integers a,b, then we get a congruence of norms b2=N(b)≡N(ε)a=1(mod⁡pn+1), so b is either 1 or -1 modulo pn+1. Hence the subgroup ε¯∩g¯⊆(OK/(pn+1))× has order 2 or 1 depending on whether there does or does not, respectively, exist an integer c such that -1≡εc(mod⁡pn+1); the existence of such a c is equivalent to the statement that 2∣r0. We will often make the following simplifying assumption.Assumption B.
Suppose thatp2∤εr0-1. (Of course, the statement in the assumption does not always hold; e.g., if p=7 and D=23·4, then ε=24+523 has order 3 modulo 7, and, in fact, 72∣ε3-1.)Making Assumption B impliesrn=pnr0 is the order of ε modulo pn+1, and thus the quotient group (OK/(pn+1))×/ε¯,g¯ is cyclic of p-prime order v=2u(p±1)/r0 where u=1 if 2∣r0 and u=0 otherwise. Choose a totally positive η∈OK whose congruence class η¯ generates this quotient. In fact, we may chooseanyη whose congruence class generates this quotient since the quantities hj,k can be modified modulo pn+1. Then we have a surjection
(54)i,j:1≤i≤pnp-1,0≤j≤v-1⟶Cm+
given by (i,j)↦[(giηj)] which is either one-to-one or two-to-one depending on whether u=0 or u=1, respectively. For each j write
(55)ηj=xj+yjδ+D2,
where xj, yj∈Z, and then define
(56)hj,k=-xjqk+yjpk
for each k. Note that χn([(giηj)])=ψn(g2i) since η has prime-to-p order modulo pn+1, so (43) and (51) imply that
(57)λpD1+λpD2=ordpn∑i=1pnp-1∑j=0v-1∑k=1rnmψn2gimmmmmmmmmmm·bk2B1hj,k-1gipn+12mmmmmmmmmmmmm-B1hj,k-1gipn+1mmmmmmmmmmmmm·B1hj,k-2gipn+1hj,k-1gipn+12.
To ease notation we define atwisted, homogeneous Dedekind sum for an arbitrary Dirichlet character ψ of modulus f:
(58)Dψa,b=∑t=1fψtatf-12btf-12.
Since the character ψn2 also generates the nth level in the cyclotomic Zp-extension of K and since t=gi runs through the units modulo pn+1 as i runs through {1,2,…,pn(p-1)}, we have proved the following.Theorem 12.
Make Assumption A. Supposep∤D is an odd prime satisfying Assumption B. Then for sufficiently large n(59)λpD1+λpD2=ordpn∑j,kbk2Dψnhj,k-1,hj,k-1-Dψnhj,k-1,hj,k-2,
where (δ+D)/2=[[b0;b1,b2,…,bm¯]] as in Lemma 2 and the rest of the notation is as above.Remark 13.
We can compute the Dedekind sumsDψn(a,a) for any integer a as follows. Write a=pma′ where p∤a′∈Z and 0≤m∈Z. If m≥n+1, then Dψn(a,a)=0 since {at/pn+1}=0 for all t∈Z. Thus we may assume 0≤m≤n. Choose b∈Z with a′b≡1(mod⁡pn+1-m). Then
(60)Dψna,a=∑t=1pn+1ψnbta′btpn+1-m-122=ψnb∑r=1p∤rpn+1-m-1rpn+1-m-122·∑q=0pm-1ψnqpn+1-m+r.
If m=0 (i.e., p∤a), then a=a′ and ab≡1(mod⁡pn+1), so
(61)Dψna,a=ψ¯na∑t=1pn+1ψntt2p2n+2
since
(62)∑t=1pn+1ψnttpn+1=∑t=1pn+1-1ψnpn+1-tpn+1-tpn+1=∑t=1pn+1-1ψnt1-tpn+1=-∑t=1pn+1ψnttpn+1=0.
On the other hand, if m>0 (i.e., p∣a), then for g∈Z a primitive root modulo all powers of p as above, we have
(63)∑q=0pm-1ψnqpn+1-m+r=∑j=0pm-1ζpni+jpn-mp-1=ζpni∑j=0pm-1ζpmjp-1=0,
where gi=r and ψn(g)=ζpn is a primitive pnth root of unity. Thus Dψn(a,a)=0, so (61) holds in this case as well since ψn(a)=0 when p∣a. We summarize the results of this remark in the following proposition.Proposition 14.
One has for alla∈Z(64)Dψna,a=ψ¯nap2n+2∑t=1pn+1-1ψnt·t2.In light of the above, one might hope to also evaluate the sums(65)Dψnhj,0,hj,-1+Dψnhj,1,hj,0+⋯+Dψnhj,rnm-1,hj,rnm-2
using a reciprocity law for Dedekind sums with characters, but the author is presently unaware of how this can be done. Nonetheless, Theorem 12 provide us with a means of computing lambda invariants.Example 15.
Takep=3 and let l≡11(mod⁡12) be a prime such that the number field K=Q(l) of discriminant D=4l has class number one. Then the totally positive fundamental unit ε>1 in K has order dividing 4 in (OK/(3))×. Suppose this order is exactly 4 and that 32∤ε4-1. Then for any positive integer n we have that 2 is a primitive root modulo 3n+1 and that (OK/(3n+1))×/ε¯,2¯ is cyclic of order 2 since the order rn of ε modulo 3n+1 will be rn=4·3n. We want a generator η∈OK of this quotient, and it clearly suffices to choose η¯ to be an element of order 8 in
(66)OK3n+1×≅Z3n⊕Z3n⊕Z8.
Alternatively, we may regard η as an eighth root of unity in Q3(l), and in that case, a fixed choice of η will suffice for all n. We can construct such an η by using Hensel lifting on 1+l since
(67)1+l4=1+l+2l2≡-l2=l≡-1mod3.
Let us consider now a concrete case. Take l=239≡11(mod⁡12). Then
(68)l=239=16,2,7,4,2,2,2,17,2,2,2,4,7,2,32¯,
so
(69)ε=p14-1+q14-1239=6195120+400729239.
It is easy to check that ε has order 4 modulo 3 since
(70)ε≡0+1239mod3
and 239≡-1(mod⁡3). We also easily verify that 9∤ε4-1 since
(71)ε4≡6+42394≡8+32392≡1+3239mod9.
For n=1, we compute the right hand side of (59) and get
(72)ordp1-12ζ3-24=ordp13ζ3-1+3=2+1=3>2=φ3,
where p1=(ζ3-1). Of course, λ3(-4)=0 since 3 remains inert in Q(i), so we must have λ3(-l)≥2 by the comments following (37). Likewise, for n=2 we get
(73)ordp272ζ325+12ζ324+72ζ323+84ζ322-12ζ32+12=6=φ32,
where p2=(ζ32-1), so λ3(-l)≥6. For n=3, we find
(74)ordp3-60ζ3317+140ζ3316+212ζ3315mnnnn+112ζ3314-40ζ3313+8ζ3311mnnnn-36ζ3310+40ζ339+184ζ337mnnnn+68ζ336-16ζ335-128ζ334mnnnn-92ζ333+136ζ332+96ζ33+36-60ζ3317+140ζ3316+212ζ3315=6<18=φ33,
where p3=(ζ33-1). Thus λ3(-239)=6. This and other similar computations (with the help of gp/pari) agree with known results as found in [17], for example.Under additional assumptions, we can compute lambda invariants without having to compute anη∈OK as above. In particular, the invariants can be computed using only a choice of primitive root g and the mod pn+1 data from the continued fraction expansion of l.Corollary 16.
Supposel≡3(mod⁡4) is a prime such that Q(l) has class number 1. For each k let pk and qk denote the numerator and denominator, respectively, of the kth convergent in the “minus” continued fraction expansion l=[[b0;b1,…,bm¯]] where m is minimal period and p-1=1, q-1=0 by convention. Let p≠l be an odd prime such that p is inert in Q(-l) and that the fundamental unit ε=pm-1+qm-1l has order r modulo p satisfying the following technical assumptions:(1)
p2∤εr-1,(2)
r=p+1 if p≡1(mod⁡4),(3)
r=(p-1)/2 if p≡3(mod⁡4).Choose g∈Z to be a primitive root modulo all powers of p, so there are integers e1, e2 with l≡ge1(mod⁡pn+1) and 2e2≡e1(mod⁡pn). For all i,k take pk(i), qk(i) to be the least nonnegative residues of gipk, giqk modulo pn+1, and let sk(i), tk(i) denote the unique integers such that bkpk-1(i)-pk-2(i)=pk(i)+sk(i)pn+1 and bkqk-1(i)-qk-2(i)=qk(i)+tk(i)pn+1. Then for sufficiently large n,
(75)λp-l+λp-4=ordpn∑i=1pnp-1ζpni∑k=1pnrmtkiqk-1i+ζpne2skipk-1i2pn+1,
where ζpn is a primitive pnth root of unity.Proof.
Obviously, Assumptions  A,  B hold forK=Q(4l) and r0=r, so we may apply all of the ideas which culminated in Theorem 12. In particular, we will exhibit a set of representatives for (OK/(pn+1))× modulo ε. The assumption that p is inert in Q(-l) is equivalent to the statement that -l is not a square modulo p. Thus for t=1,2,…,pn+1-1 with p∤t we know that tl is a unit modulo pn+1 which is never congruent to a power of ε modulo pn+1 since otherwise -t2l=N(tl)≡1(mod⁡pn+1), a contradiction. Now we use our technical assumptions on ε. In the case that p≡1(mod⁡4), we have assumed that r=p+1, so p is inert in K=Q(l) and there is a unique element of order two in (OK/(pn+1))×≅Z/(pn(p2-1))⊕Z/(pn) which corresponds to -1 and is a power of ε modulo pn+1. Consider the map
(76)t¯,tl¯:t=1,2,…,pn+1-1,p∤t⟶Cm+
given by the restriction of the map (OK/(pn+1))×→Cm+ in (44). This map is two-to-one in the case just described. Similarly, in the case that p≡3(mod⁡4), we have assumed that r=(p-1)/2 is odd, so p is split in K=Q(l) and now the map in (76) is one-to-one since -1 is not congruent to a power of ε modulo pn+1 in this case. By (43), (50), and (51), we get for n sufficiently large that λp(-l)+λp(-4) is
(77)ordpn∑t=1p∤tpn+1ψnt2ζ0,t0,tlmmmmmm+ψn-t2lζ0,tl=ordpn∑i=1pnp-1ψng2iζ0,gi+ψnge10,gilmmmmmmmmmmmmm·ζ0,gil=ordpn∑i=1pnp-1ζpni2·∑k=1pnrmbkgiqk-1pn+12-2giqk-1pn+1giqk-2pn+1mmmmm+ζpne2bkgipk-1pn+12-2gipk-1pn+1gipk-2pn+1=ordpn∑i=1pnp-1ζpni2p2n+1·∑k=1pnrmbkqk-1i-qk-2iqk-1i-qk-2iqk-1immmmm+ζpne2bkpk-1i-pk-2ipk-1i-pk-2ipk-1i=ordpn∑i=1pnp-1ζpni2p2n+1·∑k=1pnrmpn+1tkiqk-1i+qkiqk-1i-qk-1iqk-2immmmm+ζpne2pn+1skipk-1i+pkipk-1i-pk-1ipk-2i,
where pn=(1-ζpn) with ζpn=ψn2(g) and ψn(g)e1=ψn(g)2e2=ζpne2.
We have the formulaεj(pk+qkl)=pjm+k+qjm+kl for all integers j≥0, k≥-1, so the sequences pk, qk are periodic modulo pn+1 with period pnrm. Thus for all i≥0, the sequences pk(i), qk(i) are periodic with period pnrm, so
(78)∑k=1pnrmqkiqk-1i-qk-1iqk-2i=0=∑k=1pnrmpkipk-1i-pk-1ipk-2i.
The result follows.Remark 17.
Also, we note that the periodicity also implies thatpk(i), qk(i) are “palindromic” in the following sense:
(79)pk-1i=ppnrm-k-1i,qk-1i=-qpnrm-k-1i.
As remarked above, the sequence bk for k≥1 is periodic with period m and is “palindromic” with bk=bm-k for 1≤k<m while 2b0=bm=b2m=b3m=⋯. Hence if 1≤k≤pnrm/2, then bk=bpnrm-k, so
(80)pn+1ski=bkpk-1i-pk-2i-pki=bpnrm-kppnrm-k-1i-ppnrm-ki-ppnrm-k-2i=pn+1spnrm-k
and similarly tk(i)=-tpnrm-k(i). This implies that we can replace the upper index of the sum on k with pnmk/2 and still maintain the same pn-adic order. Of course, we could for the same reason ignore the 2 in the denominator of our sum in the corollary, but it is natural to include this factor of 2 since
(81)∑i=1pnp-1ζpni∑k=1pnrm/2tkiqk-1i+ζpne2skipk-1i2pn+1∈Zζpn.
In fact, since the map in (76) is two-to-one when p≡1(mod⁡4), we can replace the upper index on i with pn(p-1)/2 in this case and still conclude the sum is in Z[ζpn] with the same pn-adic order.Example 18.
Letp≥5 be a Fermat prime and let l≡3(mod⁡4) be a prime such that p is inert in Q(-l) and Q(l) has class number 1. Then l is quadratic nonresidue modulo p, so here we can choose g=l assuming additionally that p2∤lp-1-1. In this case ζpn=ψn(l) is a primitive pnth root of unity with ψn(g)2=ζpn2, so we do not have to worry about computing e2 here. For p fixed, these conditions on l are just congruence conditions modulo 4p2 plus the assumption that Q(l) has class number 1, so there should be many such examples. We just need to check the conditions on the fundamental unit in these cases in order for the corollary to apply.
Let us consider the concrete case ofp=5 and l=47. Then 5 is inert in both K=Q(l) (class number 1) and Q(-l). We have
(82)47=7;7,14¯
so m=2, the class number of Q(-l) is (4-3+17-3)/3=5, and the fundamental unit of K is
(83)ε=p2-1+q2-1l=48+747.
It is easy to check that ε has order 6=p+1 modulo 5 and that 25∤ε6-1. For n=1, we compute
(84)ordp1∑i=110ζ52i∑k=130tkiqk-1i+ζ5skipk-1i50=ordp16ζ53+ζ52+5ζ5-2=2<φ5.
Hence λ5(-47)+λ5(-4)=2. Since 5 divides the class number of Q(-47) and 5 is split in Q(i), we must have both λ5(-47)≥1 and λ5(-4)≥1, so λ5(-4)=1=λ5(-47).



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*Source: 803649-2014-11-17.xml*