A
RNAUD
J
EHANNE
M
ICHAEL
M
ÜLLER
Modularity of an odd icosahedral representation
Journal de Théorie des Nombres de Bordeaux, tome
12, n
o2 (2000),
p. 475-482
<http://www.numdam.org/item?id=JTNB_2000__12_2_475_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Modularity
of
anodd icosahedral
representation
par ARNAUD JEHANNE et MICHAELMÜLLER
d Jacques Martinet
RÉSUMÉ. Dans cet article, nour démontrons que la
représentation
03C1 de
GQ
dansd’image A5
dansPGL2(A5)
correspondant
àl’exemple
16 dans[B-K]
est modulaire. Cettereprésentation
estde conducteur 5203 et de déterminant ~-43. La modularité de
cette
représentation
n’était pas encoreprouvée ;
eneffet,
elle nevérifie pas les
hypothèses
des théorèmes de[B-D-SB-T]
et[Tay2].
ABSTRACT. In this paper, we prove that the representation p
from
GQ
in withimage A5
inPGL2(A5) corresponding
to theexample
16 in[B-K]
is modular. This representation has conductor 5203 and determinant ~-43; itsmodularity
was not yetproved,
since this representation does notsatisfy
thehypothesis
of the theorems of[B-D-SB-T]
and[Tay2].
1. Introduction Consider an irreducible Galois
representation
and the
corresponding
inducedprojective representation
where we denote
by Q
analgebraic
closureof Q
andby
GQ =Gal(Q/Q)
the absolute Galois group over
Q.
Artin has
conjectured
(for
any nontrival irreduciblerepresentation
p :GLn((C)),
that theanalytic
continuation of the L-series associated to therepresentation
p is an entire function. It is knownby
results ofHecke,
Langlands
and Tunnel([Lan], [Tun])
that p satisfies Artin’sconjecture,
ifIm(p)
is solvable. IfIm(p)
isisomorphic
toA5,
thisconjecture
is notyet
proved
ingeneral,
but it has beenproved
forparticular
casesby
Buhler([Buh])
in1978,
andby Kiming
andWang
([K-W])
in 1994. In1999,
Buz-zard,
Dickinson, Shepherd-Baron
andTaylor
proved
theconjecture
foreight
examples
[B-S]
andTaylor
gave another result that proves theconjecture
forinfinitely
many icosahedralrepresentations
([Tay2]).
In all these
examples
and theorems one has to assume, that therepre-sentation p
isodd,
thatis,
det(p)(c) =
-1,
where c denotes thecomplex
conjugation.
And for oddrepresentations,
Artin’sconjecture
isequiva-lent to the fact that the L-series of this
representation
coincides with the L-seriesL( f,
s)
of aweight
one modular formf
(see (D-S)).
We
give
one moreexample
for the Artinconjecture
withdet(p)
= X-43 and conductor 5203. With this method one canverify
Artin’sconjecture
foreach odd
representation
withquadratic
determinant and conductor lower than 10000. We prove in fact themodularity
of thisrepresentation
by
using
thecomputation
of its L-function([Jeh])
and a basis of the C-vector spaceS2(5203,1)
of cusp forms ofweight
2,
conductor 5203 andnebentype
1. Note that thisrepresentation
does notsatisfy
thehypothesis
of the maintheorem of
[B-D-SB-T].
Indeed,
if Aand /i
are theeigenvalues
of theFrobe-nius element
Frobp,2
in2,
thenA/p
has order 5. We check too that therepresentation
does notsatisfy
thehypothesis
of the theorems of[Tay2].
Acknowledgement.
It is apleasure
for the authors to thank GerhardFrey
for fruitful discussions and Kevin Buzzard for useful comments andques-tions.
2.
Modularity
of p We consider the L-functionof our irreducible two-dimensional
representation
p. We want to prove thatthe series deduced
formally
fromL(p,
s)
(where
q =e2i1rz)
is a newform ofweight
one and levelequal
to thecon-ductor of p. We make use of the
following
proposition
(see
[Frel],
proposition
1. 1).
Proposition
2.1. Assume thatbelongs
to the spaceMk(N, X)
of
modularforms of
level N andnebentype
X, where N is aninte-ger and X :
(Z/NZ) * ---+
C* a Dirichlet character. Let J-l = N+ ~)
(where
p runsthrough
the setof
prime
numbersdividing
N).
If
bn
= 0for
Let N =
8(p)
be the conductor andlet X
be the determinant of p. Via class fieldtheory,
we canconsider X,
as a Dirichlet character(7G/N7G)* ->
C* .
Proposition
2.2. be an elementsof Ml (N, X)
andIf
there are two elements g, and 92 inS2(N,
X2)
such thatand
then
belongs
to81(N, X).
Proof.
Since glbelongs
toS2(N, X)
and 0 toMl(N, X),
it isenough
to prove that themeromorphic
functiongi /0
isholomorphic.
Moreover wehave
92g2
modq¡.t/3.
. Furthermoregi
and92g2
are both elements ofS4(N,
X4),
so we canapply
proposition
2.1. This shows thatgi
= andtherefore that is
holomorphic.
DNow we consider a
projective representation
p
given
by
thepolynomial
P(X)
=X5
+5X2 -
6X + 27. It is theexample
16 of(B-K~.
We consider thelifting p
ofp
with conductor 5203 and determinant X-43(the
character associated toQ(V/’--4-3))
of[Jeh],
example
6.Theorem 2.3. The
representations
p is modular.Proof.
LetQ
be thequadratic
formQ(x, y)
= x2
+ xy +lly2;
it is known that the associated theta series/ B
belongs
toMl(43, X_43).
We recall thatf
is the series deducedformally
from
L(p, s).
For N =5203,
the element ofproposition
2.2 isequal
to 1936. We shall prove in section 3 that there are two elements gl and g2 such that
----and
By
proposition 2.2,
we conclude thatf
iscongruent
to anelement 9
=gl/8Q
in81(5203,X-43)
modulo As the seriesL(p, s)
has an Eulerqlooo
eigenvector
the associated
eigenvalue
isequal
toi 1-2v’s.
We obtain thatand then
by
proposition
2.1:It proves that there exists an
eigenform h
in81
(5203,
X-43)
witheigen-value for
T2
equal
toil-2v’5.
By
[D-S],
theorem4,
we obtain a Galois 2 *representation
associated to h. The conductor of this
representation
divides 5203 = 11 ~ .43 and its determinant isequal
to X_43· After[D-S]
theeigenvalue
ofT2
is the trace of a Frobeniusendomorphism
at 2. The tracei 1 2
can be obtainedneither
by
a tetrahedralrepresentation
(Im(ph) - A4)
norby
an octahedral oneS4).
Assume that ph is a dihedralrepresentation.
We knowthat there is a
quadratic
field and acharacter p
of the absoluteGalois group of such that ph is the induction of cp. The conductor
8(ph)
isgiven
by
the formulawhere
Od
is the discriminant of andfcp
the conductor of cp.Here,
d E{-11,-43,473}.
In
~( -43)
and inQ(v/’--11),
theprime
number 2 is inert and then ifd = -43
or -11,
the trace of theimage
of a Frobenius at 2 is zero.Hence,
we cannot have
(where
Frobp,2
is the Frobeniuselement of ph in
2).
Because of the formula(1),
theonly
ramifiedprimes
for ~p are those above 11. We denoteby p
theonly
ideal above 11 in thering
ofintegers
of~( 473).
By
[PARI],
wecompute
that the ray classgroup
of p
in~( 473)
has order 15.Hence,
theeigenvalues
of theimage
of a Frobenius
endomorphism
at 2 are 15th roots ofunity.
Therefore thetrace of a Frobenius at 2
belongs
to thecyclotomic
field which does not contain21-.
We conclude that if d =473,
we cannot have2 *
Tr(FrobPh?2)
=2 -
We have nowproved
that therepresentation
ph is icosahedral.Let
Kh
be aquintic
subfield of . Theonly possibilities
for thevaluations of the discriminant
dKh
ofKh
forprime
numbers notdividing
60 are 2 and 4(see
[Ser2],
chapter IV, §
1, 2,
3).
Since the determinantof ph is X-43, the
only possibility
for the valuation ofdKh
in 43 is 2. For the valuation ofdKh
in11,
we can have the valuations 4(if
the order ofph(hl)
isequal
to5,
whereIII
is the inertia in11),
2 or 0. We check thenin table 1 of
[B-K]
that theonly possibilities
aredKh =
112.432,
and thenph is
isomorphic
to p ordKh
=114.432,
that is toolarge
for this table.Now,
we assume thatdKh
=114.432.
In thiscase, the order
Ofph(Ill)
isequal
to 5. Since ph ismodular,
all theliftings
(where X
runsthough
the Dirichlet
characters)
are modular.Moreover,
there exists such alifting
p’
withquadratic
nontrivial determinant and such that the order ofp’ (I5 )
is 5
(see
[Serl],
theorem 5 p.228).
Sincep’
ismodular,
we can assume thatPh =
p’ (but
then,
the trace of the Frobenius in 2 of ph is no more 1but
1-2v’5,
multiplied
by
an element off 1,
-1, i,
-i}).
We can view Ph as a 2 1_ _
representation
withimage
inGL2(Z) ,
where Z is theintegral
closure of Z inQ.
We choose aplace v
in Z above 11 and we reduce ph modulo v. Weobtain a
representation
Consider the Eisenstein series
E1o
normalized such that its constant termis
equal
to 1. Then the coefhcients of theq-expansion
ofElo
are11-integers
andElo -
1 mod 11.Moreover,
the reduction 8= E
bnqn
of modulo v is a cusp form with coefhcients inIF11
ofweight
11,
level 5203 anddeterminant
jij-43
(the
reduction of X-43 modulov),
that is aneigenform
for the Hecke
operators
and such thatfor all
prime
numberp ~ {11,43}
(see
[D-S]
p. 520 and521).
Then, ;5h
satisfies Serre’sconjecture
(3.2.3?)
of[Ser3],
thatis,
Ph
arises from a cuspform with coefhcients in
IF’11.
The level of this cusp form isequal
to112.43.
In this case, we know that we can "take off" the
11-part
of this level. Thisrepresentation
arises from a cusp form with level 43(see
theorem(2.1)
p. 643 of
[Rib]).
Finally, by
theorem 4.5 p. 572 of[Edi],
we obtain that;5h
arises from a cusp form with the
weight predicted by
Serre. We concludethat
ph
satisfies Serre’sconjecture
(3.2.4?),
thatis,
with the level andweight
predicted by
Serre. Then thisrepresentation
arises from a cusp form withcoefhcients in
Fn
of level43,
characterjij-43
andweight
k as in[Ser3],
§
2.To
compute
thisweight,
we consider the restriction ofph
to the inertia group7n.
We denoteby 0
thecyclotomic
character. Since has order 5 and has a trivialdeterminant,
thenl is
conjugated
to, , , , - ,
and we obtain k = 51 or
31, using
therecipe
of[Ser3].
By
theorem 3.4 of[Edi],
that we recall below(theorem 2.4),
we see that weonly
haveto deal with the spaces of cusp forms with coefficients in
Fii
of level43,
characterX_43
andweight
3(resp. 7).
Wecompute
a basis for each ofeigenform
to
-Indeed,
we know that the values of the trace of Frobenius spanQ(i,
J5).
And the values of the reductions modulo 11 spanIF121.
Now if we lookat
S3(43, X-43),
we see that the coefficients of aneigenform
of this spacespan the root field of the
polynomial
X6
+20X4
+121X2
+214,
that iscongruent
modulo 11 to(X
+ 3) (X
+ 8) (X~ + 7X
+ 8).
Then we don’t have aneigenform
inS3(43,
k-43)
modulo 11 such that itseigenvalues
spanF121 .
For
S7(43, X-43),
the Heckeoperator
T2
has thepolynomial
characteristicThe reduction of this
polynomial
modulo 11 has two factors ofdegree
2:X 2
+ X - 3 andX 2 -
X - 3. The roots ofX 2
+ X - 3 are6 (-1
+ and6(-1 -
If the derivationapplication
acts twice on thecorresponding
eigenforms
(see
theorem 2.4below),
we obtain theeigenvalues
2(-1
+ and2(201312013B/2).
These values cannotcorrespond
with aneigenvalue -’+V’5
2
multiply by
an element off 1, -1,
i,
2013z}.
We solve the case of thepolynomial
X 2 -
X - 3by
the same way. DIn order to write theorem
2.4,
we have to introduce the derivation 0. Iff
= E
anqn
is a formalseries,
we define the seriesOf by
setting
Of
=E
Iff
is a modular form with coefficients inFp,
thenOf
is sucha modular form too.
Moreover,
theapplication 0
does neitherchange
the level nor thecharacter,
but it increases theweight by
p + 1(see
[Edi]).
Theorem 2.4
(Edixhoven).
Letf
be aneigenform
in a spaceSk(N,
X)
of
modular
forms
withcoefficients
inIE’p.
Then there existsintegers
n and k’with
0 n p - 1,
k’ p -f- 1,
and aneigenform g
inSk’
(N, X)
such thatf
and have the sameeigenvalues
for
allp).
3. Existence of gl and g2
In this section we describe very
briefly
(since
we useonly
standardtech-niques
incomputational
numbertheory)
how one canshow,
that there aretwo cusp forms gl, g2 E
S’2(5203,1)
which satisfiesand n E N as in section 2.
The method is as follows: We
compute
a basis(hi , ... ,
ofS2(5203,1)
with
-
’
(It
isenough
for us tocompute
the first 2000 coefficients of theq-expansion
of thehm ) .
Here g = dimS2 (5203,1 )
= 473 is the dimension of the space of cusp forms of level 5203. Such a basis exists due to results of[Shi],
see also
[D-D-T].
Tocompute
the basis we use the modularSymbol
method(introduced
by
Birch)
and theexplicit knowledge
of the action of the Heckealgebra
on the modularsymbols.
For adescription
of this method see forexample
[Cre]
or[Mer].
The rest is linear
algebra.
Aftercomputing
such a basis we have tocheck,
that there exists twocuspforms
gl, g2 which satisfies 2. Letbe the matrix of the coefhcients of the basis
~hl, ... ,
and the coefhcients ofthe t2
and the Then one has to show thatrank(A)
= dims2(5203,1)
= 473.(We
multiply
by
2 toget
aq-expansion
withintegral
coefficients forr~
2000.)
Tocompute
the rank we reduce the matrix A modulodiffer-ent
prime
numbers p and check thatrank(A mod p)
= 473. Hadamard’sinequality
gives
us a bound for the number ofprimes
p thatguarantee
thatrank(A)
= 473over Q
(we
need 380prime
numbers p with p >2 ’ 10~;
it isclear that these bounds are not
good,
but thecomputation
takesonly
20hours on a PII 450MHz
Linux-PC,
so we did nottry
toimprove
them).
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Arnaud JEHANNE
Laboratoire de theorie des nombres
et d’Algorithmique Arithmetique
Universite Bordeaux I
351 cours de la Liberation
33405 Talence cedex
France
E-mail : j ehanne(dmath . u-borde aux . f r
Michael MULLER
Institut fur Experimentelle Mathematik Universitat Gesamthochschule Essen Ellernstralie 29
45326 Essen
Germany