Numericalverificationofthe Stark Chinburgconjecture

English
13 Pages
Read an excerpt
Gain access to the library to view online
Learn more

Description

Numericalverificationofthe Stark-Chinburgconjecture forsomeicosahedral representations withArnaudJehanne andJonathanSands 1

  • starkprovedundersomeconditions im?¯'

  • ??? hhh

  • numericalverificationofthe stark-chinburgconjecture

  • letd?ebesuchthatan?zforall n≥1

  • k?r ???

  • ooo oo


Subjects

Informations

Published by
Reads 19
Language English
Report a problem

Numerical verification of the
Stark-Chinburg conjecture
for some icosahedral
representations
with Arnaud Jehanne
and Jonathan Sands
1The Conjecture
:G! GL (C) odd and irreducible

2
Q C
:G! C corresponding character
L(s;) corresponding Artin L-function
K
D
hi
D
D
D
for d2E, de ne
P
K\ R

f (s;) := d L(s; )

d
G


2

P
E := Q( )

s

g
sof (s;) = A n withA 2 Q for alln 1.
g
g d n n
g
g
g
g
g
g
n1
Q
Conjecture (Stark-Chinburg) Let d2E be such that A 2 Z for all
n
+
n 1, then there exists a unit "(d)2K :=K\ R such that, for all 2G:
1
0
logjj"(d) jj =f (0):
d( ()+ ())
2Present Status of the Conjecture

:G ! GL (C)! PGL (C) projective representation
2 2
Im’D (dihedral) Stark proved under some conditions
n
Im’A (tetrahedral) Chinburg numerical examples
4
Im’S (octahedral) Fogel n
4
Im’A (icosahedral)
5
14 numerical examples
3Constructing examples
^
1 !C !A !A ! 1 (non-split)
4 5 5
K
H
H
hi
H

H
H

Im’A ’ PSL (F )
H
5 2 5

2

+

^
4
K Im’A ’ ESL (F ) (determinant1)

5 2 5





hzi
4

N

K=Q a non-real quintic eld of type A
5



N=Q its Galois closure, so Gal(N=Q)’A

5

M

^


K=Q obtained by solving A !A
5 5





12






Tr



:G ! GL (C) ! C
2



F
det




:G ! GL (C) ! C (Gal(K=k) = Ker())

2









K


p


6

E := Q( ) = Q( 5;i)



O
5

k
O
O
F=Q sextic resolvent of K=Q

O
O

O
O

O
O
+

O
2 O [M : Q] = 30 and Gal(K =M)’C
4
Q
4Some reductions
p
E := Q( ) = Q( 5;i), so Gal(E=Q) =h ; i with
1 2
p p
(i) = i; ( 5) = 5;
1 1
p p
(i) =i; ( 5) = 5:
2 2
Let d2E,
X X
s
2 2
f (s) := d L(s; ) = 2< (dL(s;) +d L(s; )) = A n
d n
2 n1
with A = Tr (da ), so A 2 Z for all n 1 i
n E= n n
p p
1 i 5 + 5 5 + 5
1
d2D(E) = Z + Z + Z +i Z:
2 2 20 20
m 1
Note: "(md ) ="(d ) and"(d +d ) ="(d )"(d ) ford ;d 2D(E) , but
1 1 1 2 1 2 1 2
z
also "(id ) ="(d ) , so:
1 1
p
1 1 5+ 5
Conjecture true for all d2D(E) () Conjecture true for d = and d = .
2 20
5Computing L-functions
There exists W 2 C withjWj = 1 such that, for any u> 0:
p

X X
W N a 2n 2un
n
0
L (0;) = exp p + a Ei p
n
2 n
u N N
n1 n1
where N := conductor of and
Z
+1
t
Ei(x) = e dt=t
x
exponential integral function.
Method: compute the two sums for two di eren t values of u, deduce W (check
0
jWj = 1) and then get L (0;).
6Finding the unit
2
z z 1
Conjecture givesjj" jj for all 2G = Gal(K=Q), but ", " , " =" and
3
z z +
" =" are all real. Also, since Gal(K =M) =hzi:
2 3
z z z
P(X) = (X ")(X " )(X " )(X " )
4 3 2
=X +aX +bX +aX + 1 2O [X]:
M
Need to reconstructa and b as elements ofO with a and b known to a large
M
precision (as real numbers) and with an upper bound for the absolute value of
their conjugates over Q.
1=4 1=2
Method: Use a special quadratic form, also use " or " whenever it is
possible.
7Reconstructing an algebraic integer (1)
Let M =r (M);r (M) R and c (M);:::;c (M) C.
1 2 1 14
Problem: Find a2O such thatja j< andjr (a)j;jc (a)j<C for
M 2 j
j = 1;:::; 14.
De ne
(1) (2)
x := x; x := r (x);
2
(3) (4)
x := <(c (x)); x := =(c (x));
1 1
.............................................
(29) (30)
x := <(c (x)); x := =(c (x)):
14 14
Letf! ;! ;:::;! g integral basis of M.
1 2 30
8Reconstructing an algebraic integer (2)
31
Quadratic form on Z :

2
(1) (1)
2 2 2
Q(v ;v ;:::;v ) =C v + (C= ) v ! + +v ! v
0 1 30 1 30 0
0 1 30
30

X
2
(j) (j)
+ v ! + +v !
1 30
1 30
j=2
2
If a =a ! + +a ! is a solution, then Q(1;a ;:::;a )< 17C .
1 1 30 30 1 30
2
IfQ(v ;v ;:::;v )< 17C thenjv j 4, and if furthermorev =1, then
0 1 30 0 0
a :=v (v ! + +v ! )2O
0 1 1 30 30 M
satis es:
ja j< 4 ; jr (a)j< 4C and jc (a)j< 4C (1j 14):
2 j
9Some further veri cations
Let P(X) = Irr("~;M) and Q(X) = Irr(";~ Q).

0
Check roots of P(X) agree with exp f (0) for 2hzi.
d( ()+ ())
Check there exists an ordering"~ (2G) on the roots of Q such that

0
logjj"~ jj’f (0):
1 1

d( ( )+ ( ))
+
J=Q a degree 24 number eld such thatKJ =K , check thatKJ = Q("~).
Galois action?
10