On the p-adic Beilinson conjecture for number fields

Transcription

1 On the p-adic Beilinson conjecture for number fields Joint work with A. Besser, P. Buckingham and X.-F. Roblot Rob de Jeu jeu Department of Mathematics, Vrije Universiteit, Amsterdam On the p-adic Beilinson conjecture for number fields p. 1/27

2 The residue at s = 1 of ζ k (s) k: a number field of degree d with ring of integers O k r 1 : the number of real embeddings of k 2r 2 : the number of complex embeddings of k K 1 (O k ) = Ok has rank r = r 1 + r 2 1 σ 1,...,σ r+1 : the embeddings k C up to complex conjugation. If u 1,...,u r form a Z-basis of Ok /torsion, let 1 log σ 1 (u 1 )... log σ 1 (u r ).. R = 2r 2 [k : Q] det. 1 log σ r+1 (u 1 )... log σ r+1 (u r ) Then Res s=1 ζ k (s) = 2 r 1 (2π) r 2 R Cl(O k ) /(w D k ) [D k = discriminant of k, w = #roots of unity in k] On the p-adic Beilinson conjecture for number fields p. 2/27

3 Borel s theorem Theorem (Quillen+Borel) for n 2, K 2n 1 (k) is finitely generated; its rank m n equals r 2 for n even, r 1 + r 2 for n odd; there is a natural regulator map K 2n 1 (k) ( σ:k C R(n 1)) + = R m n, where R(m) = (2πi) m R C, and + indicates those (x σ ) σ with x σ = x σ ; the image is a lattice; if V n (k) is the volume of a fundamental domain of the image then ζ k (n) D k = q n π n(d m n) V n (k) for some q n in Q. On the p-adic Beilinson conjecture for number fields p. 3/27

4 Example If k = Q then ζ Q (s) = ζ(s) and we get n m n ζ(n) π 2 /6? π 4 /90? On the p-adic Beilinson conjecture for number fields p. 4/27

5 Interpolation formula The p-adic L-function L p of a totally real number field k satisfies L p (n,ωp 1 n,k) = E p (n,k)ζ k (n) for all integers n 0, where E p (s,k) = P p (1 N(P) s ). Here ω p : Gal(k/k) µ φ(2p) Q p is the Teichmüller character for k and p, i.e., the composition Gal(k/k) Gal(Q/Q) Gal(Q(µ 2p )/Q) = (Z/2pZ) µφ(2p) On the p-adic Beilinson conjecture for number fields p. 5/27

6 A special case of the conjecture Assume k is totally real, and take n 2 odd. Fix ordered Q-bases {a j } of k and {α j } of K 2n 1 (k) Z Q. (Both have dimension d.) Borel s map comes from composing 2 times the Beilinson regulator map reg : K 2n 1 (C) R(n 1) with σ : K 2n 1 (k) K 2n 1 (C) for all embeddings σ : k C. Let σ 1,...,σ d be the embeddings of k into C. With D 1/2, k = det(σi (a j )) and R n, (k) = det(reg σi, (α j)) Borel s theorem gives ζ k (n)d 1/2, k = q(n,k)r n, (k) for some q(n,k) in Q. On the p-adic Beilinson conjecture for number fields p. 6/27

7 Let F Q p be the topological closure of the Galois closure of k embedded into Q p in any way. There is a syntomic regulator reg p : K 2n 1 (F) F. Let σ p 1,...,σp d be the embeddings of k into F. Put D 1/2,p k = det(σ p i (a j)), R n,p (k) = det(reg p σ p i, (α j)). Conjecture (also special case of conjecture by Perrin-Riou) (1) in F we have, for some q p (n,k) in Q, L p (n,ωp 1 n,k)d 1/2,p k = q p (n,k)e p (n,k)r n,p (k) ; (2) in fact, q p (n,k) = q(n,k); (3) L p (n,ωp 1 n,k) and R n,p (k) are non-zero. On the p-adic Beilinson conjecture for number fields p. 7/27

8 A simple(?) open problem It is not known that L p (n,ωp 1 n, Q) 0 when n 2 is odd. On the p-adic Beilinson conjecture for number fields p. 8/27

9 A simple(?) open problem It is not known that L p (n,ωp 1 n, Q) 0 when n 2 is odd. For p > 2 this would certainly hold if p 1 ( 1) a (a) n = 2 1 n a=1 2 (b) n 0 in Z/pZ p 1 b=1 This seems to be the case often. Is it the case for infinitely many p when n is fixed? On the p-adic Beilinson conjecture for number fields p. 8/27

10 A motivic version Assume k/q is finite and Galois [but not necessarily totally real], with Galois group G. Fix embeddings φ : k C and φ p : k F. and R n,? (k) for? = or p are determinants of pairings D 1/2,? k and (, )? : Q[G] k C or F (σ,a) φ? (σ(a)) [, ]? : Q[G] (K 2n 1 (k) Z Q) R(n 1) or F (σ,α) reg? φ? (σ (α)) On the p-adic Beilinson conjecture for number fields p. 9/27

11 If E/Q is finite, π in E[G] an idempotent, then we can tensor k and K 2n 1 (k) with E and consider E-bilinear pairings and (, )? : E[G]π π(k Q E) E Q C or E Q F [, ]? : E[G]π π(k 2n 1 (k) Z E) E Q R(n 1) or E Q F On the p-adic Beilinson conjecture for number fields p. 10/27

12 If E/Q is finite, π in E[G] an idempotent, then we can tensor k and K 2n 1 (k) with E and consider E-bilinear pairings and (, )? : E[G]π π(k Q E) E Q C or E Q F [, ]? : E[G]π π(k 2n 1 (k) Z E) E Q R(n 1) or E Q F The dimensions for (, )? always match, for [, ]? they match in precisely two cases (for n 2): (1) the fixed field of the kernel of the representation of G on E[G]π is totally real, and n is odd; (2) the fixed field of the kernel of the representation of G on E[G]π is CM, complex conjugation of that field acts on E[G]π as multiplication by 1, and n is even. On the p-adic Beilinson conjecture for number fields p. 10/27

13 Write M E π for π(k Q E) and K 2n 1 (M E π ) for π(k 2n 1 (k) Z E). Fix ordered E-bases of those spaces, as well as of E[G]π. Let D(M E π ) 1/2,? be the determinant of the resulting pairing (, )?. In either case above, let R n,? (M E π ) be the determinant of the resulting pairing [, ]?. If χ π is the character of the representation of G on E[G]π one defines an E Q C valued L-function L(s,χ π 1, Q), with an Euler product L(s,χ π 1, Q) = p E p(s,χ π 1, Q). In the two cases above there is also a meromorphic E Q Q p -valued p-adic L-function, L p (s,χ π ωp 1 n, Q), together with an interpolation formula when s 0 is an integer. On the p-adic Beilinson conjecture for number fields p. 11/27

14 Conjecture (also...) In either case above, for n 2: (1) in E Q C we have L(n,χ π 1, Q)D(M E π ) 1/2, = e(n,m E π )R n, (M E π ) for some e(n,m E π ) in (E Q Q) ; (2) in E Q F we have L p (n,χ π ωp 1 n, Q)D(Mπ E ) 1/2,p = e p (n,m E π )E p (n,χ π 1, Q)R n,p (M E π ) for some e p (n,mπ E ) in (E Q Q) ; (3) in fact, e p (n,mπ E ) = e(n,mπ E ); (4) L p (n,χ π ωp 1 n, Q) and R n,p (Mπ E ) are units in E Q Q p and E Q F respectively. On the p-adic Beilinson conjecture for number fields p. 12/27

15 Zagier s conjecture: describing K 2n 1 (k) Let Li n (z) = j 1 zj j n (z in C with z < 1;n 0) Li 0 (z) = z/(1 z) Li 1 (z) = log(1 z) Li n+1 (z) = Li n(z)/z Li n (z) extends to a multi-valued analytic function on C \ {0, 1} on C \ {0, 1} P n (z) = π n 1 ( n 1 j=0 ) b j j! (2 log z )j Li n j (z) single-valued and satisfies P n (z) + ( 1) n P n (1/z) = 0. [b j = j-th Bernoulli number; π m = projection onto R(m) in C = R(m 1) R(m).] is On the p-adic Beilinson conjecture for number fields p. 13/27

16 For n 2: let B n (k) be a free abelian group on [x] n (x 0, 1 in k) define P n : B n (k) R(n 1) d [x] n (P n (σ(x))) σ:k C On the p-adic Beilinson conjecture for number fields p. 14/27

17 For n 2: let B n (k) be a free abelian group on [x] n (x 0, 1 in k) define P n : B n (k) R(n 1) d [x] n (P n (σ(x))) σ:k C define inductively: d n : B n (k) [x] n 2 Z k if n = 2 C n 1 (k) Z k if n > 2 { (1 x) x if n = 2 [x] n 1 x if n > 2 and C n (k) = B n (k)/ker(d n ) Ker( P n ). On the p-adic Beilinson conjecture for number fields p. 14/27

18 Conjecture (Zagier, reformulated by Deligne) For n 2: (i) there is an injection Ψ n : Ker(d n ) Ker(d n ) Ker( P n ) K 2n 1(k) Z Q with image a finitely generated group of maximal rank; (ii) Beilinson s regulator map is given by (n 1)! P n : commutes. Ker(d n ) Ker(d n ) Ker( P n ) Ψ n (n 1)! P n K 2n 1 (k) Z Q Q R(n 1) d σ reg σ On the p-adic Beilinson conjecture for number fields p. 15/27

19 Theorem (RdJ; Beilinson-Deligne) For n 2 there exists an injection Ψ n as in Zagier s conjecture such that the diagram commutes, with finitely generated image. Remark For n = 2 this was known before Zagier s conjecture and is due to Bloch and Suslin. The image in the theorem has maximal rank for: n = 2 (Suslin); n = 3 (Goncharov); all n 2 if k is cyclotomic; if ζ N = 1, ζ 1, then N[ζ] n lies in Ker(d n ) and the Ψ n (N[ζ] n ) generate K 2n 1 (k) Z Q (as Q-vector space). On the p-adic Beilinson conjecture for number fields p. 16/27

20 p-adic polylogarithms Coleman integration on P 1 C p C p = Q p p : p-adic valuation with p p = p 1 O: valuation ring F p : residue field Fix a logarithm log : C p C p such that: log(ab) = log(a) + log(b); log(1 + z) = usual power series expansion for z p small. For each x in P 1 F p (F p ) let: U x = residue disc of x = {all y in P 1 C p (C p ) that reduce to x} [a copy of the maximal ideal of O] t = t x = a local parameter on U x [e.g., t x = z x if x, t = 1/z] On the p-adic Beilinson conjecture for number fields p. 17/27

21 For x 1, in P 1 F p (F p ) let: A(U x ) = { n=0 a nt n that converge for t p < 1} A log (U x ) = A(U x ) Ω log (U x ) = A log (U x )dt For x = 1, let: A(U x ) = { n= a nt n conv. for r < t p < 1, some r < 1 } A log (U x ) = A(U x )[log t] Ω log (U x ) = A log (U x )dt Then 0 C p A log (U x ) d Ω log (U x ) 0 is exact for each x if we put dlog(t) = dt/t. On the p-adic Beilinson conjecture for number fields p. 18/27

22 Theorem (Coleman): There exists a subspace containing A Col x X(F p ) A log (U x ) A rig = lim r 1 A rig (P 1 C p \ {z such that z 1 p < r or z p > 1/r}) and such that, with Ω Col = A Col dz, 0 C p d A Col Ω Col 0 is exact. [z = affine parameter on A 1 C p ] On the p-adic Beilinson conjecture for number fields p. 19/27

23 Definition For ω in Ω Col and P,Q not in U 1 or U, let Q P ω = F ω (Q) F ω (P) for any F ω in A Col with df ω = ω. Example Put Li n+1 (z) = z 0 Li n(y)dlog y starting with Li 0 (z) = 1 z z. The Li n (z) are characterized in A Col by Li n (z) = j=1 j zj for z n p < 1 dli n+1 (z) = Li n (z)dlog(z) when n 0 Fact: the Li n (z) extend to C p \ {1}. On the p-adic Beilinson conjecture for number fields p. 20/27

24 The function P p n(z) = n 1 j=0 c j log j (z)li n j (z) with c 0 = 1 satisfies if n 1 j=0 c j (n j)! = 0. P p n(z) + ( 1) n P p n(z 1 ) = 0 We take any such function in what follows. On the p-adic Beilinson conjecture for number fields p. 21/27

25 Theorem (AB-RdJ) For σ : k F Q p let C σ n(k, O) = [x] n σ(x), 1 σ(x) are in O C n (k) = B n (k) Ker(d n ) Ker( P n ). Then Pn p,σ : B n (k) F given by [x] n Pn(σ(x)) p induces a map Pn p,σ : Cn(k, σ O) F and the solid arrows in C σ n(k, O) Ker(d n ) form a commutative diagram. Ker(d n ) Ψ n Ker(d n ) Ker( P n ) (n 1)!P p,σ (n 1)!Pn p,σ n K 2n 1 (k) Z Q F reg p σ On the p-adic Beilinson conjecture for number fields p. 22/27

26 Remark We conjecture that the dotted arrow exists and that the full diagram commutes. This holds for N[ζ] n in Ker(d n )/Ker(d n ) Ker( P n ) if ζ 1 is an N-th root of unity. On the p-adic Beilinson conjecture for number fields p. 23/27

27 Remark We conjecture that the dotted arrow exists and that the full diagram commutes. This holds for N[ζ] n in Ker(d n )/Ker(d n ) Ker( P n ) if ζ 1 is an N-th root of unity. Proposition If dim E (E[G]π) = 1 and the (motivic) conjecture applies then: (1) parts (1)-(3) hold; (2) part (4) also holds for χ π = 1, p = 2,...,19 and n = 2,...,20; (3) part (4) also holds for the 470 primitive characters χ π of Gal(Q(µ N )/Q) = (Z/NZ) with N = 3,...,49 for those values of p and n. On the p-adic Beilinson conjecture for number fields p. 23/27

28 Calculations We checked the conjecture numerically under the assumption that the dotted arrow in the last diagram exists and that the resulting diagram commutes. We checked the following cases (always for p = 2, 3, 5, 7, 11): k/q a totally real G = S 3 extension, π such that Q[G]π is an irreducible 2-dimensional representation of G (n = 3, 5) k/q a totally real G = D 8 extension, π such that Q[G]π is an irreducible 2-dimensional representation of G (n = 3, 5) k/q a CM G = D 8 extension, π such that Q[G]π is an irreducible 2-dimensional representation of G (n = 2, 4) k/q a totally real G = S 3 Z/3Z extension, π such that Q(ζ 3 )[G]π is an irreducible 2-dimensional representation of S 3 multiplied by a non-trivial character of Z/3Z (n = 3, 5) On the p-adic Beilinson conjecture for number fields p. 24/27

29 The conjecture holds numerically in all cases considered, with e(n,k)/e p (n,k) = 1 + O(p M(p) ) where M(2) = 72, M(3) = 47, M(5) = 32, M(7) = 26 and M(11) = 22 in the first three cases. It also holds in the last, where we took M(p) such that p M(p) was about On the p-adic Beilinson conjecture for number fields p. 25/27

30 Some values Splitting field of x 6 3x 5 2x 4 + 9x 3 5x + 1 G = S 3 n = 3 p R 3,p (M Q π )/D(M Q π ) 1/2,p 2 ( ) ( ) ( ) ( ) ( A928A A7888A) p L p (3,χ π ωp 2, Q) 2 ( ) ( ) ( ) ( ) (A A A278) On the p-adic Beilinson conjecture for number fields p. 26/27

31 Splitting field of x 4 2x 3 + 5x 2 4x + 2 G = D 8 n = 4 p R 4,p (M Q π )/D(M Q π ) 1/2,p 2 ( ) ( ) ( ) ( ) ( A ) p L p (4,χ π ωp 3, Q) 2 ( ) ( ) ( ) ( ) ( A2AA A7) On the p-adic Beilinson conjecture for number fields p. 27/27