The Function Θ_{K/k}(s) at s = 1

David Solomon

I. Motivation and Set-Up

• K/k finite, relatively abelian extension of number fields in ℂ • G := Gal(K/k)

• 𝔣 ⊲ O_{k} := conductor of K/k

• S_{ram} := S_{∞}(k) ∪ {𝔭 ⊲ O_{k} : 𝔭 ram. in K} = S_{∞}(k) ∪ {𝔭 : 𝔭|𝔣} • For each g ∈ G define partial zeta-function:

ζ(s; g) = ∑_{𝔞} N𝔞^{−s}, 𝔞 ⊲ O_{k}, (𝔞, 𝔣) = 1, σ_{𝔞,K} = g

for Re(s) > 1, continued to ℂ.

• Define Θ_{K/k} : ℂ ⟶ ℂG meromorphic (holomorphic at s ≠ 1):

Θ_{K/k}(s) = ∑_{g∈G} ζ(s; g)g^{−1} = ∏_{𝔭∤𝔣}(1 − N𝔭^{−s}σ_{𝔭,K}^{−1})^{−1}

(Euler product for Re(s) > 1)

• We’re considering Tate’s Θ_{K/k,S}(s) but only for S minimal, i.e. for S = S_{ram} • Properties of Θ_{K/k}(s) at s = 0: (i). (Siegel–Klingen–Shintani)

Θ_{K/k}(0) ∈ ℚG

(ii). Actually, Θ_{K/k}(0) ∈ ℚG^{−} (‘totally odd part’, except, sometimes, for an uninteresting χ_{0}-part) ⇒ little is lost by assuming K is CM, k is tot. real, so

ℚG^{−} = (1 − τ)ℚG (τ ∈ G is complex conj.)

(iii). (Deligne-Ribet, Pi. Cassou-Noguès) ann_{ℤG}(μ_{K})Θ_{K/k}(0) ⊂ ℤG (iv). Brumer’s CONJECTURE

ann_{ℤG}(μ_{K})Θ_{K/k}(0) annihilates Cl(K)

• For k = ℚ this is Stickelberger’s Theorem

• ∃ refinements/relatives of this conjecture: Conjectures of Brumer-Stark, Rubin, Popescu… Latter involve higher derivatives of Θ_{K/k} at s = 0 + ‘Stark Units’ In some cases these have been proven (Wiles, Greither, Popescu,…) and/or related to the E.T.N.C. (Burns).

• Q. Are there analogous results for Θ_{K/k}(s) at s = 1?

• If χ ∈ Ĝ then Euler product ⇒ χ(Θ_{K/k}(s)) = L_{S_{ram}}(s, χ^{−1}) (Artin/Hecke L-function with Euler factors in S_{ram} removed)

• This shows:

1) χ even ⇒ χ(Θ_{K/k}(1)) ≠ 0 (Contrast: χ(Θ_{K/k}(0)) = 0)

2) In fact, χ_{0}(Θ_{K/k}(s)) may have pole at s = 1!

• Also: Stark Conj. ⇒ very different nature of χ(Θ_{K/k}(1)) for χ odd/even

• So concentrate on odd part!

Θ_{K/k}^{−}(s) := \frac{1}{2}(1 − τ)Θ_{K/k}(s) = ∑ ; χ odd L_{S_{ram}}(s, χ^{−1})e_{χ}

(e_{χ} ∈ ℂG is idempotent)

• An Obvious Approach: Relate Θ_{K/k}^{−}(1) to Θ_{K/k}(0) χ-by-χ using F’l. Eq. of L-functions • Problem 1 Extra factors in F’l. Eq. (Gauss Sums, \sqrt{d_{k}}…)

• Problem 2 F’l. Eq. is for primitive L-functions only. imprimitive⇒ fewer Euler factors ⇒ ‘trivial’ zeroes at s = 0… … but NO trivial zeroes at s = 1!

• Nevertheless, F’l. Eq. + Siegel-Klingen-Shintani + … ⇒

\frac{1}{(πi)^{r}\sqrt{d_{k}}} Θ_{K/k}^{−}(1) ∈ ℚ(μ_{f})G

where r = [k : ℚ] and 𝔣 ∩ ℤ = fℤ.

• Q. Is there a natural analogue of D-R-CN’s integrality result?

E.g. Does there exist natural x ∈ ℤ s.t.

\frac{x}{(πi)^{r}\sqrt{d_{k}}} Θ_{K/k}^{−}(1) ∈ ℤ[μ_{f}]G?

• Problem 3 χ-by-χ method makes this hard to analyse

(More adapted to ‘… ∈ 𝔐(ℚ(μ_{f})G) ?’)

•[Historical note on Refined Stark at s = 0 vs. s = 1] • Different Approach: work g-by-g NOT χ-by-χ

• Define new function Φ_{K/k} : ℂ ⟶ ℂG replacing partial zetafunctions by twisted zeta-functions:

Φ_{K/k}(s) := (d_{k}N𝔣)^{s−1} ∑_{g∈G} Z_{K/k}(s; g)g^{−1}

(Z_{K/k}(s; g) is twisted zeta-function) The Point: •Z_{K/k}(s; g) has natural definition using additive chars. on ideals (see example below) but also …

• ∃ natural ‘Functional Equation’ in ℂG:

(∗) × Φ_{K/k}(1 − s) ↔ (∗) × Θ_{K/k}(s)

• In particular

Φ_{K/k}(0) = (−1)^{r} \frac{1}{(πi)^{r}\sqrt{d_{k}}} Θ_{K/k}^{−}(1)

(+ uninteresting correction in χ_{0}-component if k = ℚ)

• So, from now on study Φ_{K/k}(0) … • Example: k = ℚ, K = ℚ(ξ_{f}) where ξ_{f} = exp(2πi/f),

• If f ≢ 2 (mod 4) and (a, f) = 1, we find

Z_{K/k}(s; σ_{a}) := ∑_{n≥1} \frac{ξ_{f}^{an}}{n^{s}}

for Re(s) > 1, meromorphically continued

• This gives

Φ_{K/k}(0) = \frac{1}{f} ∑_{{a=1 \atop (a,f)=1}}^{f} (\frac{ξ_{f}^{a}}{1 − ξ_{f}^{a}}) σ_{a}^{−1}

• Clearly lies in ℚ(μ_{f})G! (Note: 𝔣 = fℤ)

• Possible Integrality Statement: fΦ_{K/k}(0) ∈ ℤ[μ_{f}]G if (and only if) f is not a prime power This doesn’t generalise well… • Better idea: go local

• Fix p ≠ 2 and embedding ℚ̄ ↪ ℂ_{p}

• Crucial case: k = ℚ, K = ℚ(ξ_{p^{n+1}})

• For each p-(semi)local unit u ∈ U_{p}^{1}(K) of K define the ‘ℂ_{p}Gvalued logarithm’ (a sort of logarithmic resolvent)

λ_{K/ℚ,p}(u) = ∑_{{a=1 \atop (a,p)=1}}^{p^{n+1}} log_{p}(σ_{a}u)σ_{a}^{−1}

• … and an involution ‘∗’ on ℂ_{p}G:

(∑_{g∈G} a_{g}g)^{∗} := ∑_{g∈G} a_{g}g^{−1})

Then… • Artin-Hasse’s Explicit Reciprocity Law states precisely: For all u ∈ U_{p}^{1}(K) (i). λ_{K/ℚ,p}(u)Φ_{K/ℚ}(0)^{∗} lies in ℤ_{p}G (not just in ℤ_{p}[μ_{p^{n+1}}]G)

… and if = ∑_{g∈G} c_{g}g with c_{g} ∈ ℤ_{p} then… (ii). we have the following congruence for c_{g} mod p^{n+1}:

ξ_{p^{n+1}}^{c_{g}} = ((1 − ξ_{p^{n+1}})^{−1}, gu)_{p^{n+1}} (Hilbert Symbol)

• Idea: for general K/k as above find an integrality statement for Φ_{K/k}(0) generalising (i) and …

• (If μ_{p^{n+1}} ⊂ K.) find a congruence generalising (ii)? ( Key Observation: (1 − ξ_{p^{n+1}})^{−1} ‘is’ the Stark Unit of (K^{+})^{×} ⊗ ℚ)

• Generalisation of λ_{K/ℚ,p} is an r × r, ℂ_{p}G-valued regulator R_{K/k,p} : ⋀_{ℤ_{p}G}^{r} U_{p}^{1}(K) ⟶ ℂ_{p}G

(r = [k : ℚ])

• Define a ℤ_{p}G-linear map 𝔰_{K/k} : ⋀_{ℤ_{p}G}^{r} U_{p}^{1}(K) ⟶ ℂ_{p}G

by setting

𝔰_{K/k}(u) := R_{K/k,p}(u)\sqrt{d_{k}}Φ_{K/k}(0)^{∗}

• Define 𝔖_{K/k} := image of 𝔰_{K/k} • Theorem (DS) 𝔖_{K/k} ⊂ ℚ_{p}G

• In fact, if k ≠ ℚ then ℚ_{p}𝔖_{K/k} = ℚ_{p}G^{−} (⇐ no trivial zeroes at s = 1)

• Integrality Conjecture With K/k, p as above:

𝔖_{K/k} ⊂ (∏_{{𝔭|p \atop 𝔭∤𝔣}} N𝔭)ℤ_{p}G

•Theorem (DS) Integrality Conjecture is true in the following cases 1) K/ℚ is unramified at p Pf.: Complicated expression for Φ_{K/k}(0)^{∗} in terms of the Θ_{L/k}(0) for k ⊂ L ⊂ K + D-R-CN’s result

2) p ∤ |G| Pf.: In this case

(a) 𝔐(ℚ_{p}G) = ℤ_{p}G so can work χ-by-χ.

(b) K/k is tame above p and log_{p} connects with Fröhlich’s additive theory…

3) p splits in k (+ minor restriction) Pf.: Use twisted zeta-functions directly + methods of Shintani, Coleman, Colmez.

4) k = ℚ (+ NO restriction) Pf.: explicit reciprocity laws

• Proof of (3) provides:

a) ‘formula’ for 𝔰_{K/k}(u) for u ∈ ⋀_{ℤ_{p}G}^{r} U_{p}^{1}(K) b) possible route to Integrality Conjecture (can we remove hypothesis that p splits?)

c) cases of conjecture where K/k is wildly ram. above p (see below)

• Finally: Congruences For 𝔰_{K/k}(u) We seeek generalisation of example with k = ℚ, K = ℚ(ξ_{p^{n+1}}) Use Hilbert symbols and Rubin-Stark units.

• Assume:

1) Integrality Conjecture is satisfied for K/k (⇒ 𝔰_{K/k}(u) ∈ ℤ_{p}G ∀u)

2) μ_{p^{n+1}} ⊂ K

3) |S_{ram}| ≥ r + 1 (e.g. k fixed, n → ∞)

• K^{+} := max. real subfield of K G^{+} := Gal(K^{+}/k)

• Complex Rubin-Stark conjecture for K^{+}/k implies existence of ε^{+} ∈ ℚ ⊗ ⋀_{ℤG^{+}}^{r} K^{+,×} s.t.

Θ_{K^{+}/k,S_{ram}}(s) = R_{K^{+}/k,∞}(ε^{+})s^{r} + o(s^{r}) as s → 0 where R_{K^{+}/k,∞} is archimedean regulator: ℚ⊗⋀_{ℤG^{+}}^{r} K^{+,×} ⟶ ℝG^{+} • Extra condits. make ε^{+} unique as elt. of ℚ ⊗ ⋀_{ℤG^{+}}^{r} K^{+,×}

• Let ε := image of ε^{+} in ℚ ⊗ ⋀_{ℤG}^{r} K^{×}

• Example: k = ℚ, K = ℚ(ξ_{p^{n+1}}) as before:

ε^{+} = −\frac{1}{2} ⊗ (1 − ξ_{p^{n+1}})(1 − ξ_{p^{n+1}}^{−1}) ∈ ℚ ⊗ K^{+,×}

ε = 1 ⊗ (1 − ξ_{p^{n+1}})^{−1} ∈ ℚ ⊗ K^{×} (∈ 1 ⊗ K^{×})

• For simplicity suppose ε ∈ 1 ⊗ K^{×} also for our extension K/k (maybe always true in our situation?)

• Define a pairing H_{n} : ⋀_{ℤG}^{r} K^{×} × ⋀_{ℤ_{p}G}^{r} U_{p}^{1}(K) ⟶ (ℤ/p^{n+1}ℤ)G

(Determinant of a matrix over (ℤ/p^{n+1}ℤ)G defined by the distinct Hilbert symbols above p. Details omitted)

• Congruence Conjecture With these hypotheses ∃ ε̃ ∈ ⋀_{ℤG}^{r} K^{×} s.t. ε = 1 ⊗ ε̃ and

𝔰_{K/k}(u) ≡ H_{n}(ε̃, u) (mod p^{n+1}) ∀u ∈ ⋀_{ℤ_{p}G}^{r} U_{p}^{1}(K)

• Evidence for Congruence Conjecture:

1) Holds if k = ℚ

2) Suppose: (i). Rubin-Stark Conj. and its p-adic analogue hold with a certain

strong norm coherence in cyclotomic ℤ_{p}-tower over K^{+} (ii). K/k, p satisfy certain conditions, notably p splits in k Then I can deduce a slightly weakened form of Congruence Conjecture by ‘p-adic analytic continuation’ of p-adic Rubin-Stark Conjecture from s = 1 to s = 0, explicit reciprocity law etc.

• Note: conditions of (2) are satisfied for large class of cases with K/ℚ abelian but k ≠ ℚ.