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Numerical verification of theStark-Chinburg conjecture forsome icosahedral representations

with Arnaud Jehanne
and Jonathan Sands&%
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'$The Conjecture¯Q⊂CK DD hτ iDD

K∩RGQggggggggg ΓE := Q(ψ)ρ : G → GL2 (C)odd and irreducible
ψ:G→C
corresponding character
L(s, ρ)corresponding Artin L-function
for d ∈ E, define
P γ
fd (s, ρ) :=d L(s, ργ )γ∈ΓP−sso fd (s, ρ) =An n with An ∈ Q for all n ≥ 1.
n≥1Conjecture (Stark-Chinburg) Let d ∈ E be such that An ∈ Z for all n ≥ 1, thenthere exists a unit ε(d) ∈ K ∩ R such that, for all σ ∈ G:−10log ||ε(d)σ || = fd(ψ(σ)+ψ(στ)) (0).

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Present Status of the Conjectureρρ̄ : G −→ GL2 (C) → PGL2 (C)Imρ̄ ' Dn (dihedral)projective representation

Stark (1980)

proved under some conditions

Imρ̄ ' A4 (tetrahedral) Chinburg (1980)

5 numerical examples

Imρ̄ ' S4 (octahedral)

8 numerical examples

Fogel (1998)

Imρ̄ ' A5 (icosahedral)
14 numerical examples&%3'$Constructing examples
K LLL hτ iLLL
L2

4K∩R
4 hziN
KF
12FK
6k OOOOO

OOO
2 OOO

5Q1 −→ C4 −→ Â5 −→ A5 −→ 1 (non-split)Imρ̄ ' A5 ' PSL2 (F5 )Imρ ' Â5 ' ESL2 (F5 ) (determinant ±1)
K/Q a non-real quintic field of type A5
N/Q its Galois closure, so Gal(N/Q) ' A5
K/Q obtained by solving Â5 → A5ρTrρdet

ψ : G −→ GL2 (C) −→ Cχ : G −→ GL2 (C) −→ C× (Gal(K/k) = Ker(χ))√E := Q(ψ) = Q( 5, i)
F/Q sextic resolvent of K/Q[KF : Q] = 30 and Gal(K ∩ R/KF ) ' C4&%
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'$Some reductions√
E := Q(ψ) = Q( 5, i), so Gal(E/Q) = hγ1 , γ2 i with
γ1 (i) = −i,
γ2 (i) = i,√√γ1 ( 5) = 5,√√γ2 ( 5) = − 5.

Let d ∈ E,
fd (s) :=Xγγγ2γ2d L(s, ρ ) = 2< (dL(s, ρ) + d L(s, ρ )) =XAn n−sn≥1γ∈Γwith An = TrE/Q (dan ), so An ∈ Z for all n ≥ 1 iff√√i5+ 55+ 51d ∈ D(E)−1 = Z + Z +Z+iZ.
222020Note: ε(m d1 ) = ε(d1 )m and ε(d1 + d2 ) = ε(d1 ) × ε(d2 ) for d1 , d2 ∈ D(E)−1 , but alsoε(id1 ) = ε(d1 )z , so:−1

Conjecture true for all d ∈ D(E)√5+ 51Conjecture true for d = 2 and d = 20 .

⇐⇒&%
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Computing L-functions

There exists W ∈ C with |W | = 1 such that, for any u > 0:
√ XXan
2πn2πunW NL0 (0, ρ) =exp − √+an E i √2π n≥1 nu N
Nn≥1
where N := conductor of ρ and Ei (x) =R +∞ −t
e dt/t exponential integral function.
xMethod: compute the two sums for two different values of u, deduce W (check|W | = 1) and then get L0 (0, ρ).

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'$Finding the unit

23

Conjecture gives ||εσ || for all σ ∈ G = Gal(K/Q), but ε, εz , εz = ε−1 and εz = ε−z
are all real. Also, since Gal(K ∩ R/KF ) = hzi:
23P (X) = (X − ε)(X − εz )(X − εz )(X − εz )= X 4 + aX 3 + bX 2 + aX + 1∈ OKF [X].

Need to reconstruct a and b as elements of OKF with a and b known to a large
precision (as real numbers) and with an upper bound for the absolute value of theirconjugates over Q.

Method: Use a special quadratic form, also use ε1/4 or ε1/2 whenever it is possible.

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Reconstructing an algebraic integer
Let M = KF : M = r1 (M ), r2 (M ) ⊂ R and c1 (M ), . . . , c14 (M ) ⊂ C.
Problem: Find a ∈ OM such that |a − α| <  and |r2 (a)|, |cj (a)| < C for j = 1, . . . , 14.
Define x(1) := x, x(2) := r2 (x), x(j) := <(cl−2 (x)) + =(cl−2 (x)) (3 ≤ j ≤ 16) andx(j) := <(cl−16 (x)) − =(cl−16 (x)) (17 ≤ j ≤ 30).
Let OM = Zω1 + Zω2 + · · · + Zω30 and for set
Q(v0 , v1 , . . . , v30 ) = C 2 v02 + (C/)2

(1)(1)
v1 ω1 + · · · + v30 ω30 − v0 α+30 X

2(j)(j)v1 ω1 + · · · + v30 ω30

2j=2• If a = a1 ω1 + · · · + a30 ω30 is a solution, then Q(1, a1 , . . . , a30 ) < 31C 2 .
• If Q(v0 , v1 , . . . , v3 ) < 31C 2 then |v0 | ≤ 5, and if furthermore v0 = ±1, thena = v0 (v1 ω1 + · · · + v30 ω30 ) satisfies:√√√|a − α| < 30, |r2 (a)| < 30C and |cj (a)| ≤ 15C (1 ≤ j ≤ 14).
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Some further verifications

Let P (X) = Irr(ε̃, KF ) and Q(X) = Irr(ε̃, Q).
0• Check the roots of P (X) agree with exp fd(ψ(σ)+ψ(στ )) (0) for σ ∈ hzi.
• Check there exists an ordering ε̃σ (σ ∈ G) on the roots of Q such that0log ||ε̃σ || ' fd(ψ(σ−1 )+ψ(σ −1 τ )) (0).

• Given J/Q a degree 24 number field such that KJ = K ∩ R, check that
KJ = Q(ε̃).
• Galois action?

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'$The examples
NdkKpower+twist?

1948 = 22 · 487−487

X 5 − 7X 3 − 17X 2 + 18X + 73

1/4

No2083−2083

X 5 + 8X 3 + 7X 2 + 172X + 53

1/4

No

2336 = 25 · 73

−22 · 73

X 5 + 2X 3 − 4X 2 − 2X + 4

1/2

χ−42707−2707

X 5 − 2X 4 + 2X 3 − 8X 2 + 21X − 62

1/4

No2863 = 7 · 409−7 · 409

X 5 − 2X 4 + 6X 3 + X 2 + 14X + 49

1/4

χ−73004 = 22 · 751−751

X 5 − 8X 3 + 10X 2 + 160X + 128

1/2

No3203−3203

X 5 − X 4 − X 3 − 9X 2 + 20X − 111/4

No3547−3547

X 5 − 8X 3 − 2X 2 + 31X + 74

1/4

No3548 = 22 · 887−887

X 5 + 10X 3 + 10X 2 + 44X + 56

1/2

No

3587 = 17 · 211−17 · 211

X 5 − 2X 4 − 3X 3 − 13X 2 + 51X − 171/4

χ173676 = 22 · 919−919

X 5 − 8X 3 + 28X 2 − 40X + 48

1/2

No&%
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√

An example: N = 7 · 409, d = (5 + 5)/20X

120+ 26X

119− 94X− 3470848290X110118117− 30730083146X− 663689188701923X− 487528951945104522X− 13600X

99104− 281311X

109− 2021804070374122X− 913545334089156134170X−34980994234137834004988972X

8682−364225668247380528514085586X− 8612234037145139046881703X−578852895069490862382330272X−742769471405376488763892122X− 325889924154119365692755X

20984+ 1983824193814413999886771962X

4751605754

31− 51289870206760479995338X− 2697514791586967527902X

27− 913545334089156134170X

26242322− 633454733X

814− 3783005384356929126X− 4006528158838154X− 4241975499068X− 154147794X

713650837165625956
53−1081353097405617020513141512X

42−220610572593979107867316098X− 17205248221768863457571082X

34− 836465690752396946216978X− 20456291234678745439466X− 294293366619039740386X− 487528951945104522X
175− 281311X

113413733292116− 3470848290X− 13600X

494525− 663689188701923X− 30730083146X
47568383079+914406880475444830539894282X

12− 3321996X

87

46− 2021804070374122X− 442488173187X− 26695600X&18

91+914406880475444830539894282X+1775424928237598246145749032X− 127082471589904301709030X

197295−1081353097405617020513141512X+ 2187960822311042822094511360X

28− 4884414082098948X

76+ 3760223467003604255883124473X− 1996086627778243950531172X− 20403327017395397024X

105

100

−220610572593979107867316098X+ 10499011099495497101938960668X−1096185494282559030925178618X

111− 294293366619039740386X+ 8564829773982177534419666602X

35

32

96

+ 2372728685456958572223960652X− 34980994234137834004988972X− 4292965781360113755325812X− 36581684427924657X+ 2425873616614254748075556939X39− 69290016038721990395949600X− 160018034060140X− 17205248221768863457571082X

80−364225668247380528514085586X− 29683203312874X− 1715520846X

66
63− 1715520846X− 836465690752396946216978X

43− 84955151957338226945X

1555+ 6039722171959017044736565850X+2320289738861916855748839350X− 7663680434531394529531X− 36581684427924657X+ 6039722171959017044736565850X

69+ 11217034388314946629995466683X

58

88−15033934417574738736045691X+ 1983824193814413999886771962X−578852895069490862382330272X

3673

112

− 20456291234678745439466X−850511718430215025448509179X+ 2320289738861916855748839350X−742769471405376488763892122X

40− 8612234037145139046881703X

676461+ 2372728685456958572223960652X
5270

92

−128003082848154913592158939X

77106101− 84955151957338226945X

85+ 8564829773982177534419666602X

44− 4884414082098948X− 4292965781360113755325812X

81− 633454733X− 29683203312874X− 325889924154119365692755X+ 10499011099495497101938960668X− 128003082848154913592158939X− 20403327017395397024X

97

107

113

89

+ 3760223467003604255883124473X−850511718430215025448509179X− 4006528158838154X

102− 154147794X− 127082471589904301709030X+ 2187960822311042822094511360X−15033934417574738736045691X

114− 7663680434531394529531X

74

48− 26695600X− 4241975499068X− 2697514791586967527902X+ 1775424928237598246145749032X+2425873616614254748075556939X

98

108

−69290016038721990395949600X

78−1096185494282559030925178618X

103

115

93

90

− 51289870206760479995338X− 1996086627778243950531172X

94− 3321996X− 442488173187X− 3783005384356929126X− 160018034060140X

116− 94X

210+ 26X + 1%
11