Computing complete lists of Artin representations given a group, character, and conductor bound

Transcription

1 Computing complete lists of Artin representations given a group, character, and conductor bound John Jones Joint work with David Roberts 7/27/15 Computing Artin Representations John Jones SoMSS 1 / 21

2 Dedekind ζ-function If K is a number field, ζ K (s) = P ( 1 N(P) s ) 1 If K/Q is abelian with Galois group G, ζ K (s) = χ Ĝ L(s, χ) Would like to factor ζ K (s) for more general K Computing Artin Representations John Jones SoMSS 2 / 21

3 Complex representations let G be a finite group only finitely many irreducible complex representations of G every complex representation equivalent to a sum of irreducibles, unique up to order if G is abelian, then these are the Dirirchlet characters associated to G If K/Q is Galois and G = Gal(K/Q) Define Artin L-function for a representation ρ : G Aut(C n ) (ignoring bad factors) L(s, ρ) = char ρ(frob(p)) (p s ) 1 p where char A (t) = det(i ta)) Computing Artin Representations John Jones SoMSS 3 / 21

4 Factoring Let K be a number field, n = [K : Q] G = Gal(K gal /Q) get σ : G S n, permutation representation induces complex representation ρ on space of complex functions on {1,..., n}. ρ = i ν i with ν i irreducible. Then ζ K (s) = L(s, ρ) = i L(s, ν i ) Computing Artin Representations John Jones SoMSS 4 / 21

5 Conductor Given G = Gal(K gal /Q), ρ : G Aut(C n ) Exists conductor f ρ Z + Define locally using higher ramification groups f ρ1 ρ 2 = f ρ1 f ρ2 If ρ comes from permutation representation of K/Q, f ρ = Disc(K) 1/ deg(ρ) Sometimes useful to think in terms of root conductor: fρ Computing Artin Representations John Jones SoMSS 5 / 21

6 What s our problem? Given Given a finite group G χ an irreducible complex character of G of a faithful representation a bound B Can we compute all Galois extensions K/Q with Gal(K/Q) = G with Artin representation ρ whose character is χ and f χ B? Hope to use large enough bounds to find first examples, if not lists of the first few examples. Computing Artin Representations John Jones SoMSS 6 / 21

7 Simplification Note Galois conjugate characters have the same conductor. If a character is not defined over Q, take its trace: it has the same root conductor. The resulting representation may not be defined over Q, but a multiple of it (Schur index) is, so again the root conductor is the same. Result: suffices to study irreducible rational representations, and we get information about Galois conjugate characters together. Computing Artin Representations John Jones SoMSS 7 / 21

8 Well-posedness problem When we want to connect χ and G with Gal(K/Q), we are picking an isomorphism G = Gal(K/Q). Different choices differ by composition with some ψ Aut(G). If ρ : G Aut(V) is a complex representation, ψ Aut(G), then ρ ψ is another representation with the same conductor. Starting with G, characters which differ by ψ Aut(G) are indistinguishable with respect to possible Artin conductors. So we group these as well. Computing Artin Representations John Jones SoMSS 8 / 21

9 Plan Given G, χ, and B 1 Relate bound B to a bound for a number field search 2 Do number field search 3 Compute f χ for each field to select winners 4 Bask in the fame and fortune which follows from a successful number theory computation Computing Artin Representations John Jones SoMSS 9 / 21

10 Step 1: connect to number field search Given G, χ, and B, which fields should we look at? Trivial cases: when our search is already a standard number field search. G = S 3 has characters χ 1a = 1, χ 1b, χ 2 (1 = trivial character, subscripts are degree and a label as needed). Cubic S 3 field K has permutation character 1 + χ 2, so f χ2 = D K Similarly for G = C 3 which has complex characters 1, χ 1b, and χ 1c = χ 1b : D K = f 1b f 1c = f 2 1b Similarly any C p with p prime Many cases do not fit this situation, starting with degree 4 Galois groups. Computing Artin Representations John Jones SoMSS 10 / 21

11 Step 1 generic cases We use the tame-wild principle: to figure out a divisibility relation between conductors, just consider tame cases and it will work for all Computing Artin Representations John Jones SoMSS 11 / 21

12 Step 1 generic cases We use the tame-wild principle: to figure out a divisibility relation between conductors, just consider tame cases and it will work for all Does not work in complete generality, but Theorem (JR) f χ f α(χ) r(g) where r(g) is the regular representation and α(χ) is the tame constant. α(χ) is a purely group theoretic constant which is easy to compute Computing Artin Representations John Jones SoMSS 11 / 21

13 Applying tame-wild If we have K such that f χ B, D K gal α(χ) = f α(χ) r(g) so the field is found by our search. f χ B So, given G, χ, and B, compute all K satisfying D K gal B 1/α(χ). Computing Artin Representations John Jones SoMSS 12 / 21

14 Applying tame-wild If we have K such that f χ B, D K gal α(χ) = f α(χ) r(g) so the field is found by our search. f χ B So, given G, χ, and B, compute all K satisfying D K gal B 1/α(χ). To compute α(χ), we need to know how to compute conductors in tame cases. Computing Artin Representations John Jones SoMSS 12 / 21

15 Tame conductors Given a representation ρ of G and an element g G such that g = I p, v p (f χ ) = #eigenvalues λ of ρ(g) s.t. λ 1 counting multiplicities. Computing Artin Representations John Jones SoMSS 13 / 21

16 Tame conductors Given a representation ρ of G and an element g G such that g = I p, v p (f χ ) = #eigenvalues λ of ρ(g) s.t. λ 1 counting multiplicities. Clearly additive and f ρ1 ρ 2 = f ρ1 f ρ2 Computing Artin Representations John Jones SoMSS 13 / 21

17 Tame conductors Given a representation ρ of G and an element g G such that g = I p, counting multiplicities. v p (f χ ) = #eigenvalues λ of ρ(g) s.t. λ 1 Clearly additive and f ρ1 ρ 2 = f ρ1 f ρ2 In a permutation representation, ρ(g) with cycle type e 1 e k, we want k i=1 e i 1 an m-cycle in a permutation representation has eigenvalues: each m-th root of unity, exactly once a local tame totally ramified extension of degree e i has discriminant p ei 1 the e i from the cycle type are the ramification indices Computing Artin Representations John Jones SoMSS 13 / 21

18 More tame conductors ζ Write A = ρ(g) = , m = multiple of order of A. 0 0 z k m ζ i = i=1 { 0 ζ 1 m ζ = 1 Tr(A + A A m )/m = multiplicity of 1 as eigenvalue Can compute this for a (rational) character from a (rational) character table Computing Artin Representations John Jones SoMSS 14 / 21

19 Step 2: Computing number fields Want to find all minimal degree stem fields K with Gal(K gal /Q) = G and Disc(K gal ) B This is (by far) the most time consuming part of the process For solvable G: class field theory many cases (base field/modulus/congruence subgroup) may have to approach K by computing some other stem field first K K = rd(k) rd(l) For nonsolvable G: targeted Hunter search many small searches of tiny targets a target is a discriminant coming from a particular ramification pattern different ramification patterns may have different effects on Disc(K gal ) e.g., tame e i patterns 221 and 311 contribute p 2 to Disc(K), but p 1/2 and p 2/3 to root discriminant of K gal Computing Artin Representations John Jones SoMSS 15 / 21

20 Step 3: Computing conductors Need to compute f χ for each field to cull out winners Use Magma programs due to Tim Dokchitser works locally at each ramifying prime computes local splitting field Work globally Connect conductor to discriminants of resolvent fields for a permutation character, conductor = Disc for the resolvent field By Artin induction theorem, every rational valued character is a rational linear combination of permutation characters Determining these combinations is row reduction on a small matrix of rationals (rational character values) Discriminants of resolvent polynomials may be huge Doesn t matter we already know the ramifying primes Computing Artin Representations John Jones SoMSS 16 / 21

21 S 5 Character table on left, tame conductor exponents on right. λ α(χ) χ 1a χ 1b χ 4a χ 4b χ 5a χ 5b χ r(s 5 ) α(χ) is normalized for root conductor. Case giving minimum ratio in bold. Computing Artin Representations John Jones SoMSS 17 / 21

22 More S 5 Computed all 1,351 quintic S 5 fields with D K gal : α(χ) 75 α(χ) # Min root Pos χ 4a χ 4b = 122, χ 5a = 1,778, χ 5b = 380, χ = 36,081, Entries with correspond to number fields of one higher degree. root = root conductor Pos = ranking in full list by discriminant of Galois closure Computing Artin Representations John Jones SoMSS 18 / 21

23 More S 5 Computed all 1,351 quintic S 5 fields with D K gal : α(χ) 75 α(χ) # Min root Pos χ 4a χ 4b = 122, χ 5a = 1,778, χ 5b = 380, χ = 36,081, Entries with correspond to number fields of one higher degree. root = root conductor Pos = ranking in full list by discriminant of Galois closure Artin induction: f 4a = D 5 f 4b = D 10 D 1 5 D 1 2 f 5a = D 6 f 5b = D 12 D 1 6 D 1 2 f 6 = D 30 D 6 D 2 2D 2 5 D 2 12 Computing Artin Representations John Jones SoMSS 18 / 21

24 Other results G χ f 1/ deg # C 2 1b C 3 1b S C 4 1b D A S 4 3a b C 5 1b D F A G χ f 1/ deg # S 5 4a b a b C 6 1b D S 3 C A 4 C S 3 S C3 2 C 4 4a, b S 4 C 2 3a, b C3 2 D 4 4a, b c, d Computing Artin Representations John Jones SoMSS 19 / 21

25 Other results G χ f 1/ deg # A 6 5a, b S 6 5a, b c, d a b a, b C 7 1b D C 7 C F G χ f 1/ deg # GL 3 (2) A a b S 7 6a b a Computing Artin Representations John Jones SoMSS 20 / 21

26 The end Thank you. Computing Artin Representations John Jones SoMSS 21 / 21