computing complete lists of artin representations given a...

Background Basic Question Step 1 Step 2 Step 3 Results

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

John Jones

Joint work with David Roberts

7/27/15

Computing Artin Representations John Jones SoMSS 1 / 21


Dedekind ζ-function

If K is a number field,

ζK(s) =∏

P

(1− N(P)−s)−1

If K/Q is abelian with Galois group G,

ζK(s) =∏χ∈G

L(s, χ)

Would like to factor ζK(s) for more general K



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, uniqueup 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(Cn) (ignoringbad factors)

L(s, ρ) =∏

p

charρ(Frob(p))(p−s)−1

wherecharA(t) = det(I − tA))



Factoring

Let

K be a number field, n = [K : Q]

G = Gal(Kgal/Q)

get σ : G→ Sn, permutation representation

induces complex representation ρ on space of complex functions on{1, . . . , n}.ρ = ⊕iνi with νi irreducible.

ThenζK(s) = L(s, ρ) =

∏i

L(s, νi)



Conductor

Given G = Gal(Kgal/Q), ρ : G→ Aut(Cn)

Exists conductor fρ ∈ Z+

Define locally using higher ramification groups

fρ1⊕ρ2 = fρ1 · fρ2

If ρ comes from permutation representation of K/Q,

fρ = |Disc(K)|

Sometimes useful to think in terms of root conductor: f 1/ deg(ρ)ρ



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 Artinrepresentation ρ whose character is χ and fχ ≤ B?

Hope to use large enough bounds to find first examples, if not lists of the firstfew examples.



Simplification

Note

Galois conjugate characters have the same conductor.

If a character is not defined over Q, take its trace: it has the same rootconductor.

The resulting representation may not be defined over Q, but a multiple ofit (Schur index) is, so again the root conductor is the same.

Result: suffices to study irreducible rational representations, and we getinformation about Galois conjugate characters together.



Well-posedness problem

When we want to connect χ and G with Gal(K/Q), we are picking anisomorphism G ∼= Gal(K/Q). Different choices differ by composition withsome ψ ∈ Aut(G).

If ρ : G→ Aut(V) is a complex representation, ψ ∈ Aut(G), then ρ ◦ ψ isanother representation with the same conductor.

Starting with G, characters which differ by ψ ∈ Aut(G) are indistinguishablewith respect to possible Artin conductors. So we group these as well.



Plan

Given G, χ, and B1 Relate bound B to a bound for a number field search2 Do number field search3 Compute fχ for each field to select winners4 Bask in the fame and fortune which follows from a successful number

theory computation



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 = S3 has characters χ1a = 1, χ1b, χ2(1 = trivial character, subscripts are degree and a label as needed).

Cubic S3 field K has permutation character 1 + χ2, so fχ2 = |DK |Similarly for G = C3 which has complex characters 1, χ1b, andχ1c = χ1b: |DK | = f1b · f1c = f 2

1b

Similarly any Cp with p prime

Many cases do not fit this situation, starting with degree 4 Galois groups.



Step 1 – generic cases

We use the tame-wild principle: to figure out a divisibility relationbetween conductors, just consider tame cases and it will work for all



Step 1 – generic cases

We use the tame-wild principle: to figure out a divisibility relationbetween conductors, just consider tame cases and it will work for all

Does not work in complete generality, but

Theorem (JR)

fχ ≥ 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



Applying tame-wild

If we have K such that fχ ≤ B,

|DKgal |α(χ) = f α(χ)r(G) ≤ fχ ≤ B

so the field is found by our search.

So, given G, χ, and B, compute all K satisfying |DKgal | ≤ B1/α(χ).



Applying tame-wild

If we have K such that fχ ≤ B,

|DKgal |α(χ) = f α(χ)r(G) ≤ fχ ≤ B

so the field is found by our search.

So, given G, χ, and B, compute all K satisfying |DKgal | ≤ B1/α(χ).

To compute α(χ), we need to know how to compute conductors in tamecases.



Tame conductors

Given a representation ρ of G and an element g ∈ G such that 〈g〉 = Ip,

vp(fχ) = #eigenvalues λ of ρ(g) s.t. λ 6= 1

counting multiplicities.



Tame conductors




Clearly additive and fρ1⊕ρ2 = fρ1 · fρ2



Tame conductors




Clearly additive and fρ1⊕ρ2 = fρ1 · fρ2

In a permutation representation, ρ(g) with cycle type e1 · · · ek, we want∑ki=1 ei − 1

an m-cycle in a permutation representation has eigenvalues: each m-th rootof unity, exactly oncea local tame totally ramified extension of degree ei has discriminant pei−1

the ei from the cycle type are the ramification indices



More tame conductors

Write A = ρ(g) =

ζ1 0 0

0. . . 0

0 0 zk

, m = multiple of order of A.

m∑i=1

ζ i =

{0 ζ 6= 1m ζ = 1

Tr(A + A2 + · · ·+ Am)/m = multiplicity of 1 as eigenvalue

Can compute this for a (rational) character from a (rational) character table



Step 2: Computing number fields

Want to find all minimal degree stem fields K with Gal(Kgal/Q) ∼= Gand |Disc(Kgal)| ≤ B

This is (by far) the most time consuming part of the process

For solvable G: class field theorymany cases (base field/modulus/congruence subgroup)may have to approach K by computing some other stem field firstK ⊆ K =⇒ rd(K) ≤ rd(L)

For nonsolvable G: targeted Hunter searchmany small searches of tiny targetsa target is a discriminant coming from a particular ramification patterndifferent ramification patterns may have different effects on Disc(Kgal)e.g., tame ei patterns 221 and 311 contribute p2 to Disc(K), but p1/2 andp2/3 to root discriminant of Kgal



Step 3: Computing conductors

Need to compute fχ for each field to cull out winnersUse Magma

programs due to Tim Dokchitserworks locally at each ramifying primecomputes local splitting field

Work globallyConnect conductor to discriminants of resolvent fieldsfor a permutation character, conductor = |Disc| for the resolvent fieldBy Artin induction theorem, every rational valued character is a rationallinear combination of permutation charactersDetermining these combinations is row reduction on a small matrix ofrationals (rational character values)Discriminants of resolvent polynomials may be hugeDoesn’t matter – we already know the ramifying primes



S5

Character table on left, tame conductor exponents on right.

λ5 15 221 312 5 213 41 32 15 221 312 5 213 41 32 α(χ)

χ1a 1 1 1 1 1 1 1 0 0 0 0 0 0 0χ1b 1 1 1 1 −1 −1 −1 0 0 0 0 1 1 1χ4a 4 0 1 −1 2 0 −1 0 2 2 4 1 3 3 0.50χ4b 4 0 1 −1 −2 0 1 0 2 2 4 3 3 3 0.75χ5a 5 1 −1 0 1 −1 1 0 2 4 4 2 4 4 0.80χ5b 5 1 −1 0 −1 1 −1 0 2 4 4 3 3 5 0.80χ6 6 −2 0 1 0 0 0 0 4 4 4 3 5 5 0.83

r(S5) 120 0 0 0 0 0 0 0 60 80 96 60 90 100

α(χ) is normalized for root conductor. Case giving minimum ratio in bold.



More S5

Computed all 1,351 quintic S5 fields with |DKgal | ≤ 75120:

α(χ) 75α(χ) # Min root Posχ4a∗ .50 8.66 33 1609 6.33 103χ4b .75 25.48 7 52173 = 122,825 18.72 21χ5a∗ .80 31.62 116 263473 = 1,778,112 17.78 116χ5b .80 31.62 357 2432892 = 380,208 16.27 6χ6 .83 36.52 73 2433174 = 36,081,072 18.18 12

Entries with ∗ correspond to number fields of one higher degree.root = root conductorPos = ranking in full list by discriminant of Galois closure



More S5

Computed all 1,351 quintic S5 fields with |DKgal | ≤ 75120:

α(χ) 75α(χ) # Min root Posχ4a∗ .50 8.66 33 1609 6.33 103χ4b .75 25.48 7 52173 = 122,825 18.72 21χ5a∗ .80 31.62 116 263473 = 1,778,112 17.78 116χ5b .80 31.62 357 2432892 = 380,208 16.27 6χ6 .83 36.52 73 2433174 = 36,081,072 18.18 12

Entries with ∗ correspond to number fields of one higher degree.root = root conductorPos = ranking in full list by discriminant of Galois closure

Artin induction: f4a = |D5| f4b = |D10D−15 D−1

2 | f5a = |D6|

f5b = |D12D−16 D−1

2 | f6 = |D30D6D22D−2

5 D−212 |



Other results

G χ f 1/ deg #

C2 1b 3.00 6086C3 1b 7.00 451S3 2 4.80 1886C4 1b 5.00 489D4 2 6.24 9868A4 3 14.64 270S4 3a 11.30 779

3b 6.12 1603C5 1b 11.00 49D5 2 6.86 736F5 4 13.69 470A5 3 26.45 12

4 11.66 325 12.35 161

G χ f 1/ deg #

S5 4a 6.33 334b 18.72 75a 16.27 3575b 17.78 1166 18.18 73

C6 1b 7.00 402D6 2 9.33 10242

S3 × C3 2 7.21 275A4 × C2 2 8.60 98S3 × S3 4 14.83 132C2

3 o C4 4a, b 17.80 13S4 × C2 3a, b 6.92 3694C2

3 o D4 4a, b 7.9 84c, d 23.36 24



Other results

G χ f 1/ deg #

A6 5a, b 12.35 08 42.81 09 28.20 210 30.61 0

S6 5a, b 11.53 05c, d 6.82 09a 21.10 09b 31.88 0

10a, b 24.22 016 35.46 0

C7 1b 29.00 15D7 2 8.43 464

C7 o C3 3 34.93 8F7 6 18.34 95

G χ f 1/ deg #

GL3(2) 3 26.06 16 11.23 17 32.44 08 23.16 1

A7 6 12.54 010 41.21 014a 37.95 014b 39.43 015 35.73 021 38.33 035 41.28 0

S7 6a 7.55 06b 29.98 014a 31.43 0



The end

Thank you.


computing complete lists of artin representations given a...

Documents