Numerical verification of theStark-Chinburg conjecture
for some icosahedral
representations

with Arnaud Jehanne
and Jonathan Sands

Q̄⊂CKDiDDhτ
DK∩RGgggggggg ΓThe Conjecture

E := Q(ψ)

Qg

ρ : G → GL2 (C)odd and irreducible
ψ:G→C
corresponding characterL(s, ρ)corresponding Artin L-function
for d ∈ E, defineP
fd (s, ρ) :=dγ L(s, ργ )γ∈ΓP
so fd (s, ρ) =n≥1

An n−s with An ∈ Q for all n ≥ 1.

Conjecture (Stark-Chinburg) Let d ∈ E be such that An ∈ Z for alln ≥ 1, then there exists a unit ε(d) ∈ K+ := K ∩ R such that, for all σ ∈ G:
0log ||ε(d)σ || = fd(ψ(σ)+ψ(στ)) (0).
−1

Present Status of the Conjectureρρ̄ : G −→ GL2 (C) → PGL2 (C)Imρ̄ ' Dn (dihedral)StarkImρ̄ ' A4 (tetrahedral) ChinburgImρ̄ ' S4 (octahedral)

Fogel

projective representation

proved under some conditions
numerical examples
numerical examples

Imρ̄ ' A5 (icosahedral)
14 numerical examples

Constructing examples

KH i HHHhτHH
2 4K+4 hzi
 N
M

12

FK6
5k OOO
OOOOO
2 OOO Q1 −→ 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 → A5det

ρ

Tr

ρ

ψ : G −→ GL2 (C) −→ C

χ : G −→ GL2 (C) −→ C× (Gal(K/k) = Ker(χ))√E := Q(ψ) = Q( 5, i)
F/Q sextic resolvent of K/Q[M : Q] = 30 and Gal(K+ /M ) ' C4

Some reductions√
E := Q(ψ) = Q( 5, i), so Gal(E/Q) = hγ1 , γ2 i with√√
γ1 (i) = −i, γ1 ( 5) = 5,√√γ2 ( 5) = − 5.
γ2 (i) = i,Let d ∈ E,
fd (s) :=X

dγ L(s, ργ ) = 2< (dL(s, ρ) + dγ2 L(s, ργ2 )) =XAn n−sn≥1γ∈Γwith An = TrE/ (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, butalso ε(id1) = ε(d1 )z , so:√Conjecture true for all d ∈ D(E)−1 ⇐⇒ Conjecture true for d = 21 and d = 5+20 5 .

Computing L-functions

There exists W ∈ C with |W | = 1 such that, for any u > 0:√X
2πunW N Xa
2πnnan Ei √exp − √+L0 (0, ρ) =2π n≥1 nu NNn≥1
where N := conductor of ρ andEi(x) =Z +∞e−t dt/tx

exponential integral function.

Method: compute the two sums for two different values of u, deduce W (check|W | = 1) and then get L0 (0, ρ).

Finding the unit

2

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

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

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

Reconstructing an algebraic integer (1)Let 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.

Definex(4):= <(c1 (x)),x(3)x(2):= x,x(1):= r2 (x),:= =(c1 (x)),.............................................
x(29) := <(c14 (x)),x(30) := =(c14 (x)).

Let {ω1 , ω2 , . . . , ω30 } integral basis of M .

Reconstructing an algebraic integer (2)Quadratic form on Z31 :2(1)(1)
Q(v0 , v1 , . . . , v30 ) = C 2 v02 + (C/)2 v1 ω1 + · · · + v30 ω30 − v0 α+30 
Xj=2(j)(j)v1 ω1 + · · · + v30 ω30

2• If a = a1 ω1 + · · · + a30 ω30 is a solution, then Q(1, a1 , . . . , a30 ) < 17C 2 .
• If Q(v0 , v1 , . . . , v30 ) < 17C 2 then |v0 | ≤ 4, and if furthermore v0 = ±1, thena := v0 (v1 ω1 + · · · + v30 ω30 ) ∈ OM
satisfies:|a − α| < 4,|r2 (a)| < 4C

and |cj (a)| < 4C (1 ≤ j ≤ 14).

Some further verifications

Let P (X) = Irr(ε̃, M ) and Q(X) = Irr(ε̃, Q).
0• Check 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).

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

The abelian condition

K
K+ (ε1/4)

K+tt
tt
tt
tt

4 or 84

K+ (ε1/4 )/M is an abelianextensionof group C4 or C8M
KQ

The examples

X 5 + 8X 3 + 7X 2 + 172X + 53−2083

X 5 − 7X 3 − 17X 2 + 18X + 73

−487

1948 = 22 · 487

KdkN

2083

−2707

2707

−22 · 73

2336 = 25 · 73

X 5 + 2X 3 − 4X 2 − 2X + 4

X 5 − 2X 4 + 2X 3 − 8X 2 + 21X − 62

X 5 − X 4 − X 3 − 9X 2 + 20X − 11−3203

3203−751

X 5 − 2X 4 + 6X 3 + X 2 + 14X + 49−7 · 409

3004 = 22 · 751

2863 = 7 · 409

3547−3547−17 · 211

3587 = 17 · 211−887

3548 = 22 · 887
3676 = 22 · 919−919

X 5 − 8X 3 + 10X 2 + 160X + 128
X 5 − 8X 3 − 2X 2 + 31X + 74

X 5 + 10X 3 + 10X 2 + 44X + 56
X 5 − 2X 4 − 3X 3 − 13X 2 + 51X − 17X 5 − 8X 3 + 28X 2 − 40X + 48

power
1/4
1/4
1/2
1/4
1/4
1/2
1/4
1/4
1/2
1/4
1/2+twist?
NoNo
χ−4No
χ−7NoNoNoNo
χ17No

X

120+ 26X

119− 94X

118− 3470848290X− 13600X11011799104

N = 7 · 409, d = (5 +− 281311X− 30730083146X− 663689188701923X− 487528951945104522X

109116− 3321996X− 442488173187X− 2021804070374122X− 3783005384356929126X− 913545334089156134170X− 51289870206760479995338X− 1996086627778243950531172X− 34980994234137834004988972X− 364225668247380528514085586X

94

90868278− 1096185494282559030925178618X

524810398115

1085)/20114102− 154147794X
10781139213718612− 633454733X

5101106112− 1715520846X− 160018034060140X− 36581684427924657X
96111

1051009587
837975716865625956
53+ 914406880475444830539894282X− 1081353097405617020513141512X− 220610572593979107867316098X− 17205248221768863457571082X− 836465690752396946216978X− 20456291234678745439466X− 294293366619039740386X− 487528951945104522X
1711− 281311X

44945413733292521− 663689188701923X− 30730083146X− 3321996X

91− 836465690752396946216978X− 17205248221768863457571082X− 220610572593979107867316098X+ 914406880475444830539894282X

50

4642383430

2622− 2021804070374122X− 442488173187X− 26695600X− 1081353097405617020513141512X

72+ 2425873616614254748075556939X+ 2372728685456958572223960652X+ 8564829773982177534419666602X+ 10499011099495497101938960668X+ 3760223467003604255883124473X+ 2187960822311042822094511360X− 51289870206760479995338X− 913545334089156134170X− 3783005384356929126X− 154147794X− 294293366619039740386X− 20456291234678745439466X
88848076− 15033934417574738736045691X
6966
6360

5754

− 34980994234137834004988972X− 1996086627778243950531172X

312723− 4006528158838154X− 4241975499068X− 633454733X− 29683203312874X− 4884414082098948X

97− 84955151957338226945X− 7663680434531394529531X− 325889924154119365692755X− 8612234037145139046881703X− 128003082848154913592158939X− 850511718430215025448509179X

73+ 2320289738861916855748839350X+ 1983824193814413999886771962X+ 6039722171959017044736565850X+ 6039722171959017044736565850X+ 1983824193814413999886771962X− 2697514791586967527902X− 20403327017395397024X
19− 364225668247380528514085586X

3935− 127082471589904301709030X

2824

14+ 1775424928237598246145749032X− 1096185494282559030925178618X

43− 4292965781360113755325812X

3251

47− 578852895069490862382330272X

40− 69290016038721990395949600X

3620− 4884414082098948X− 29683203312874X
9√− 26695600X− 4241975499068X− 4006528158838154X

93898581
77− 742769471405376488763892122X
70676461+ 11217034388314946629995466683X

5855+ 2320289738861916855748839350X− 742769471405376488763892122X

4415− 20403327017395397024X− 2697514791586967527902X− 127082471589904301709030X− 4292965781360113755325812X− 69290016038721990395949600X− 578852895069490862382330272X

74+ 1775424928237598246145749032X+ 2187960822311042822094511360X+ 3760223467003604255883124473X+ 10499011099495497101938960668X+ 8564829773982177534419666602X+ 2372728685456958572223960652X+ 2425873616614254748075556939X− 15033934417574738736045691X− 850511718430215025448509179X− 128003082848154913592158939X− 8612234037145139046881703X− 325889924154119365692755X− 7663680434531394529531X− 84955151957338226945X− 36581684427924657X− 160018034060140X− 1715520846X

16− 3470848290X− 13600X

310− 94X

2+ 26X + 1