Explicit units and a refinement of the Rubin-Stark conjecture

Anthony Hayward King’s College, London

http://www.mth.kcl.ac.uk/∼hayward 1. Abelian Stark conjecture

Let L/K be an abelian extension of global fields with group G. Let S be a finite nonempty set of primes of K containing all archimedean and ramified primes. Let U_{K,S} = S-units of K, U_{L,S} similarly for L. Let w_{L}, w_{K} be the numbers of roots of unity.

Suppose S contains one place v which splits completely in L/K. Let w be a place of L above v.

Abelian Stark Conjecture (Tate’s form) There is ε ∈ U_{L,S} such that

1) w_{L} · ζ′_{L/K,S}(0, σ) = − ln | ε^{σ} |_{w} ∀σ ∈ G;

2) L(ε^{1/w_{L}}) is abelian over K.

Example K = ℚ, L = ℚ(ζ_{m})^{+}, S = Supp(m)∪ {∞}, ε = (−ζ_{m}^{−1})(1 − ζ_{m})^{2}.

Rubin reformulated Stark’s conjecture: let T be a finite set of places of K disjoint from S such that U_{L,S,T} (=S-units of L congruent to 1 modulo every place of L above a place in T) is torsion free. I.e. if w_{L,T} = number of roots of unity in U_{L,S,T} then w_{L,T} = 1. Set δ_{T} = ∏_{v∈T} (1 − N_{v}σ_{v}^{−1}).

Abelian Stark Conjecture (Rubin’s reformulation) For (almost) all T = {v}, there is u ∈ U_{L,S,T} such that ζ′_{L/K,S}(0, σ)−N_{v}ζ′_{L/K,S}(0, σ_{v}^{−1}σ) = − ln | u^{σ} |_{w}.

Connection: u = (ε^{1/w_{L}})^{δ_{T}}. Note: also formulated for any number of splitting primes.

Advantage of Rubin’s form: no w_{L}. Fix K, S, T, let L vary. For number fields, if L/K unramified above p and p−1 > [K : ℚ] then p ∤ w_{L} and, for any v with residue characteristic p, w_{L,v} = 1. So such a T = {v} works for all L simultaneously.

Conjecture determines u uniquely. So get system of special units over K satisfying Euler system distribution relation. Get norm-compatible systems u_{S} ∈ \underleftarrow{lim}_{L} U_{L,S,T}, distribution relation under norm operator

\underleftarrow{lim}_{L} U_{L,S∪\{v′\},T} → \underleftarrow{lim}_{L} U_{L,S,T} …

2. Explicit units

Let K be a global field with ≤ 1 archimedean place. Let ∞ be the archimedean place (or any place for K a function field). Let O = the ∞- integers of K.

For any ideal 𝔪 of O, have ray class field mod 𝔪 (totally split at ∞) H(𝔪). This corresponds to the idèle subgroup K^{×}U(𝔪), where idèles in U(𝔪) are local units away from ∞ and congruent to 1 (mod 𝔪).

Fix 𝔪 = 𝔭_{1}^{e_{1}}𝔭_{2}^{e_{2}} ⋯ 𝔭_{n}^{e_{n}}. For a subset I ⊆ {1, …, n} define 𝔪_{I} = ∏_{i∈I} 𝔭_{i}^{e_{i}}. Abbreviate w_{I} := w_{K,𝔪_{I}}.

Fix an embedding ∞ : H(𝔪) ↪ ˆ \overline{K̂}.

Proposition There exists a system λ̂_{I} ∈ K^{ab} for I ⊆ {1, …, n} such that for 𝔱 coprime to 3𝔪w_{H(𝔪)}, λ_{I} := λ̂_{I}^{δ_{𝔱}} satisfies

1) λ_{I} ∈ U_{H(𝔪_{I}),S_{I},𝔱};

2a) N_{H(𝔪_{J})/H(𝔪_{I})}(λ_{J}^{w_{I}/w_{J}}) = λ_{I}^{∏_{J−I}(1−σ_{pi}^{−1})};

2b) N_{H(𝔪_{i})/H}(λ_{i}^{w_{K}/w_{i}})O_{H} = 𝔭_{i}^{1−N_{𝔱}}O_{H} (O_{H} = ring of integers of Hilbert class field);

3) w_{I}(ζ′_{H(𝔪_{I})/K,S_{I}}(0, σ)−N_{v}ζ′_{H(𝔪_{I})/K,S_{I}}(0, σ_{𝔱}^{−1}σ))

= − ln | λ_{I}^{σ} |_{∞} .

Remark: 1) (3) ⟹ (2a). Does it imply (2b)?!

2) This verifies abelian Rubin-Stark conjecture in these contexts via embedding L into a H(𝔪) for sufficiently large 𝔪 with w_{K,𝔪} = 1.

Remarks on the proof

For K = ℚ, we take λ̂_{I} = (1 − ζ_{m_{I}}), so that λ_{I} = (1 − ζ_{m_{I}})^{δ_{𝔱}} = \frac{(1−ζ)}{(1−ζ^{t−1})^{t}}.

If K is a function field, we use sign-normalised rank-one Drinfeld modules ρ : O → ℂ_{∞}[T]_{∘}. The 𝔪_{I}-torsion ρ[𝔪_{I}] ⊆ ℂ_{∞} is naturally an Omodule isomorphic to O/𝔪_{I}. We choose λ̂_{I} to be an O-generator of this module. Then λ̂_{I} generates a “narrow” ray class field and λ_{I} satisfies all the properties of the Proposition.

For K an imaginary quadratic field, we consider the function defined relative to an inclusion of lattices L ⊆ L′:

F (z, L, L′) = \sqrt[12]{\frac{Δ(L)^{N_{𝔱}}}{Δ(L′)}} ∏_{u} (℘_{L}(z) − ℘_{L}(u))^{−1}
= \frac{ϕ(z, ω)^{N_{𝔱}}}{\sqrt[12]{1} · ϕ(z, ω′)}

where u runs through (L′/L)/{±1} − {0}, ω = (w_{1}, w_{2}) with L = ℤw_{1} + ℤw_{2}, the twelfth roots are carefully chosen to ensure certain distribution relations, and ϕ is the function in Kronecker’s second limit formula. We let L = 𝔪_{I} and L′ = 𝔱^{−1}𝔪_{I}. Then most of the properties are known to hold for λ_{I} = F (1, 𝔪_{I}, 𝔱^{−1}𝔪_{I}) (the last is the limit formula). To see the congruence mod 𝔱, note that ϕ(1, ω) ∈ H(12f_{𝔪_{I}}^{2}). Can show that ϕ(1, ω)^{σ_{𝔱}} = \sqrt[12]{1} · ϕ(1, ω′). This shows that F (1, 𝔪_{I}, 𝔱^{−1}𝔪_{I}) = ϕ(1, ω)^{N_{𝔱}−σ_{𝔱}}.

3. A refinement of the conjecture

Back to setting of the abelian Stark conjecture (L/K abelian, S containing one place which splits completely).

Let ϕ ∈ Hom_{ℤ[G]}(U_{L,S,T} , ℤ[G]). What is the image ϕ(u) ∈ ℤ[G]? When there are no splitting places, Gross’s conjecture postulates a congruence for

Θ_{L/K,S,T} (0) := δ_{T} ∑_{σ∈G} ζ_{L/K,S}(0, σ)σ^{−1} ∈ ℤ[G];

it should be ≡ h_{K,S,T} Reg (mod I^{#S}), where I = augmentation ideal and Reg is an “algebraic regulator”. Burns has refined Rubin’s conjecture to accommodate this idea. The formulation originates in explicit-unit interpretations of the equivariant Tamagawa number conjecture.

In the one-place-splitting case, Burns’s conjecture says: (set n := #S − 1)

Conjecture ϕ(u) ≡ h_{K,S,T} Reg^{ϕ} (mod I^{n}) where

Reg^{ϕ} = | \begin{matrix}ϕ^{1}(u_{1}) & ⋯ & ϕ^{1}(u_{n}) \\ f_{v_{1}}(u_{1}) − 1 & ⋯ & f_{v_{1}}(u_{n}) − 1 \\ ⋮ & ... & ⋮ \\ f_{v_{n−1}}(u_{1}) − 1 & ⋯ & f_{v_{n−1}}(u_{n}) − 1 \end{matrix} |

=∑_{i=1}^{n}(−1)^{i+1}ϕ^{1}(u_{i}) \underbrace{| \begin{matrix}f_{v1}(u_{1})−1 & ⋯ & f_{v1}(u_{n})−1 \\ ⋮ & ... & ⋮ \\ f_{vn−1}(u_{1})−1 & ⋯ & f_{vn−1}(u_{n})−1 \end{matrix} |}_{excluding u_{i}}

• ϕ^{1}(x) ∈ ℤ is the coefficient of 1 ∈ G in ϕ(x);

• u_{j} is a basis of U_{K,S,T};

• v_{i} are the nonsplit places;

• f_{v} : K_{v}^{×} → G is the local reciprocity map for L/K localised at v.

Remarks

• Formulated for any number of places splitting.

• Proved for extensions of degree two.

• Provides interpretation for “minus-unit” conjectures of Darmon and Gross.

• Generalised version deduced from ETNC in many cases (recent work of Burns). Question: how much can be proved using the explicit units of §2 in the case where K has ≤ 1 archimedean place, and ∞ splits completely in L?

4. Congruence for explicit units

Let K be a global field with at most one infinite place, 𝔪 an ideal of O, S = Supp(𝔪) ∪ {∞}, T = {𝔱}.

Theorem Let b be an integer such that w_{K} | b and there is a section to the restriction map Gal(H(𝔪)/K) → Gal(H/K) whose b-th power is multiplicative. Then Burns’s conjecture is true after multiplication by b on both sides.

Sketch-proof: Step 1: using the distribution relations, produce an intermediate congruence involving σ_{𝔭_{i}}.

Step 2: Relate the global symbols to the local ones using product formula.

Step 1 involves some interesting combinatorics. The proof is by induction on #S, using a combinatorial identity concerning minors of matrices with zero row sum – cf. work of Greither and Kuc̆era.

Fix a_{ij} in some ring for 1 ≤ i ≠ j ≤ n. For I ⊆ {1, …, n} let A^{I} be the #I × #I matrix indexed by I with off-diagonal entries a_{ij} and zero row-sum. For J ⊆ {1, …, n}, let A_{J}^{I} be the determinant of the minor matrix given by deleting the rows and columns indexed by I ∩ J in A^{I}. The Matrix Tree Theorem shows how to express A_{J}^{I} in terms of graph-theoretic “arborescences”.

Matrix Tree Theorem (S. Chaiken and D. J. Kleitman) A_{J}^{I} = (−1)^{#I−#J} ∑_{T} A(T), where T runs through all arborescences on I with root J.

Example

A_{\{3,4\}}^{\{1,2,3,4\}} = det ( \begin{matrix}−a_{12} − a_{13} − a_{14} & a_{12} \\ a_{21} & −a_{21} − a_{23} − a_{24} \end{matrix} )

= a_{13}a_{21} + a_{12}a_{23} + a_{13}a_{23} + a_{13}a_{24} + a_{14}a_{23} + a_{14}a_{24} + a_{12}a_{24} + a_{14}a_{21}, corresponding to the arborescences

\overbrace{\begin{matrix}2 \\ ↓ \\ 1 \\ ↓ \\ 3 & 4 \end{matrix}}^{a_{13}a_{21}}  \overbrace{\begin{matrix}1 \\ ↓ \\ 2 \\ ↓ \\ 3 & 4 \end{matrix}}^{a_{12}a_{23}}  \overbrace{\begin{matrix}1 & 2 \\ ↓ & ↙ \\ 3 & 4 \end{matrix}}^{a_{13}a_{23}}  \overbrace{\begin{matrix}1 & 2 \\ ↓ & ↓ \\ 3 & 4 \end{matrix}}^{a_{13}a_{24}}  \overbrace{\begin{matrix}2 & 1 \\ ↓ & ↓ \\ 3 & 4 \end{matrix}}^{a_{14}a_{23}}  \overbrace{\begin{matrix}1 & 2 \\ ↘ & ↓ \\ 3 & 4 \end{matrix}}^{a_{14}a_{24}}  \overbrace{\begin{aligned}& 1 \\ & ↓ \\ & 2 \\ & ↓ \\ 3 & 4 \end{aligned}}^{a_{12}a_{24}}  \overbrace{\begin{aligned}& 2 \\ & ↓ \\ & 1 \\ & ↓ \\ 3 & 4 \end{aligned}}^{a_{14}a_{21}}

The identity we need is, for fixed J ⊆ {1, …, n}:

∑_{I}(−1)^{#I′∩J}F (#I)A_{J}^{I} ∏_{i∈I′-J}(∑_{j∈I} a_{ij}) = 0,

where F (X) = (1 − X)^{#J−1}. By interpreting this as counting “leaves” on “trees”, reduce this expression to

± ∑_{T} A(T) ∑_{r=0}^{m}(−1)^{r}\binom{m}{r}F (n − r)

where m = number of leaves on arborescence T (vertices {1, …, n}, root J). This is then zero by Theorem (Euler) If 0 ≤ t < m,

∑_{r=0}^{m}(−1)^{r}\binom{m}{r}r^{t} = 0.