LXXXII.1(1997)
The spa e of period polynomials
by
Shinji Fukuhara (Tokyo)
1. Introdu tion. The purpose of this note is to enhan e our under-
standing of the spa e of period polynomials. The period polynomials has
been studied in onne tion with modular integrals (e.g., Knopp [4℄), usp
forms via the Ei hler{Shimuraisomorphism(e.g., Kohnen{Zagier [5℄), and
withvariousothertopi sofmathemati s(Zagier[8℄). LetK bea eld,and
X be an indeterminate. Let V
w
=V
w
(K) denote thespa e of polynomials
in X of degree w (even positive)with oeÆ ients inK. Then V
w (K) is
an(w+1)-dimensionalve torspa e,whi hmaybeidenti edwiththespa e
L
w
n=0 K(X
n
). LetG=GL
2
(Z)=1. Then Ga ts on V
w via
(1:1) (Pj )(X)=P
aX+b
X+d
( X+d) w
for =
a b
d
2 G and P(X) 2 V
w
. Now we de ne the spa e, W
w , of
period polynomials of weight w to be the subspa e of V
w
hara terized by
the followingproperties: W
w
=ker(1+S)\ker(1+U +U 2
) (see [1, 4, 5,
7℄), namely,
W
w
=fP 2V
w
:P +PjS=P +PjU +PjU 2
=0g
whereS =
0 1
1 0
and U =
1 1
1 0
.
Though the spa e W
w
is interesting in its own right, it might be more
naturalto onsiderperiodLaurentpolynomials,ratherthanperiodpolyno-
mials alone. Let b
V
w
be the spa e K(X 1
)V
w
K(X w+1
). The spa e,
W
w
, of period Laurent polynomials an be de ned in a similar way, and it
turnsoutto bea subspa eof b
V
w
. InLemma 2.2, itwillbeshown thatW
w
1991 Mathemati sSubje tClassi ation: Primary11F20,11F11;Se ondary11B68.
Keywordsandphrases: periodpolynomial, uspform,modularform,Ei hler{Shimura
isomorphism.
TheauthorwishestothankProfessorN.Yuiforherusefuladvi e.
is a odimensionone subspa eof
W
w
. From the de nition, learly1 X w
belongsto W
w
anditrepresentsa \trivial"element. Therefore, itmightbe
morenaturalto onsiderthequotientspa esW
w
=h1 X w
iand
W
w
=h1 X w
i.
The main purpose of this note is to onstru t homomorphisms q :
V
w
! b
V
w
=h1 X w
i and q :V
w
! b
V
w
=h1 X w
i, anddes ribe theirimages
expli itly. (See Theorem 3.3, Lemmas 4.1 and 5.1.) The two mappings
q and q areexpressed usingBernoullipolynomials,and their imagesare
indeedidenti edwith
W
w
=h1 X w
i. Consequently,thiswillyieldaspanning
set fq (X n
)g w
n=0
or fq (X n
)g w
n=0 of
W
w
=h1 X w
i. We obtain a relation
betweenthepolynomials (X n
)andr
(R
n
)ofKohnen{Zagier[5℄,andthus,
we show that
(1:2) fr
(R
n )g
w
n=0
mod1 X w
spans W
w
=h1 X w
i:
In Kohnen{Zagier[5, p.203℄, thefa t (1.2) wasobtainedusingtheisomor-
phismtheoremofEi hlerandShimuraforperiodmappings. Inthisnote,we
take a reverse route fromthat of [5℄, namely,we rst onstru t a spanning
set forthe spa eW
w
=h1 X w
i in terms of thehomomorphismsq , and as
a orollaryof thisresult,we redis overtheisomorphismtheorem ofEi hler
and Shimuraforperiodmappings.
2. Preliminaries. Throughoutthepaperwe assumethat w is aneven
positiveinteger. Forea h w,let
V
w
(K)=fpolynomials ofdegree winX with oeÆ ients inKg:
We oftenwriteV
w forV
w
(K)when the eldK isplain. Thea tionofGon
V
w
de nedin(1.1) an beextended to an a tionofthe groupring ZGby
P
X
n
i
i
= X
n
i (Pj
i )
for n
i
2 Z and
i
2 G. Using the a tion, a subspa e W
w of V
w
an be
des ribedas
W
w
=ker(1+S)\ker (1+U +U 2
) (2:1)
=fP 2V :P +PjS=P +PjU+PjU 2
=0g (2:2)
forS =
0 1
1 0
andU =
1 1
1 0
. Following Kohnen{Zagier[5, p. 199℄, we
onsiderthea tionofa spe i element,i.e., "=
1 0
0 1
. Itisshownin[5℄
thatWj"=W and there is adire tsum de omposition
W
w
=W +
w
W
w
of W
w
su hthat Pj"=P forP 2W
w
. Morepre isely,
W +
w
=fP 2W
w
:P is aneven polynomialg;
W =fP 2W
w
:P is anodd polynomialg:
We mayalso onsiderthespa e
b
V = n
n
X
i=m
i X
i
:m;n2Zsu h thatmn;
i 2K
o
of Laurentpolynomialsinone variable and itssubspa e
b
V
w
= n
w+1
X
i= 1
i X
i
:
i 2K
o
:
For 2GandP 2 b
V,Pj isde nedby(1.1). Itisnolongeranelement
of b
V, but a rational fun tion. However, the equation Pj = 0 will make
sense. The spa e
W
w
is de nedsimilarlyto (2.2), i.e.,
W
w
=fP 2 b
V : P +PjS=P +PjU +PjU 2
=0g:
ClearlyW
w
W
w
. Moreover,
W
w
are de nedsimilarlyto W
w :
W +
w
=fP 2
W
w
:P is an even Laurentpolynomialg;
W
w
=fP 2
W
w
:P is an odd Laurentpolynomialg:
Itisobviousthat
W
w
=
W +
w
W
w
. We allanelementofW
w
(respe tively,
W
w
)aperiod polynomial (respe tively,period Laurentpolynomial)ofweight
w. We also all W
w
(respe tively,
W
w
) the spa e of period polynomials
(respe tively,period Laurent polynomials) ofweight w.
Now we onsider a spe ialelement of b
V
w
. Let f
w
be an element of b
V
w
de nedby
f
w (X)=
w+2
X
n=0
neven B
n B
w+2 n
n!(w+2 n)!
X n 1
:
It wasshown inZagier[7, p.453℄that f
w 2
W
w
(the homogeneous version
of this fa t was also proved in [3℄). Let hf
w
i denote the subspa e of
W
w
whi h is spanned by f
w
. We are interested in how di erent
W
w
and W
w
are. Thiswillbe answered inLemmas 2.2 and 2.3. We willalso showthat
W
w
b
V
w .
Lemma 2.1. For m 2, let P(X) = P
w+m
i= m
i X
i
bea Laurent polyno-
mial su h that Pj(1+S)=Pj(1+U +U 2
)=0. Then
m
=
w+m
=0 .
Proof. Sin e Pj(1+S)=0,wehave
w+m
X
i X
i
+ w+m
X
i
1
X
i
X w
=0;
namely ,
(2:3)
i
+( 1) w i
w i
=0 fori= m;:::;0;:::;w+m:
Sin e Pj(1+U+U 2
)=0,wehave
(2:4) w+m
X
i= m
i X
i
+ w+m
X
i= m
i
X 1
X
i
X w
+ w+m
X
i= m
i
1
X 1
i
(X 1) w
=0:
Multiply(2.4) byX m
(X 1) m
to obtain
(2:5)
w+m
X
i= m
i X
m+i
(X 1) m
+ w+m
X
i= m
i
(X 1) m+i
X w i+m
+ w+m
X
i= m
i ( 1)
i
(X 1) w i+m
X m
=0:
We al ulatethe oeÆ ientofX 1
onthelefthandsideof(2.5). Sin em2
bytheassumption,we obtain
(2:6) ( 1) m 1
m
m
+( 1)(w+2m)
w+m
+( 1) m
m+1
+( 1)
w+m 1
=0:
Theequations(2.6)and (2.3)implythat(w+m)
w+m
=0. Hen e
w+m
=
0,and then
m
=0by(2.3), ompletingthe proof.
Lemma 2.2.
W
w
=W
w
hf
w i:
Proof. Let P(X) = P
w+m
i= m
i X
i
belong to
W
w
. Then we an assume
m=1 by Lemma2.1above. Set
Q(X)=P(X)
1
(w+2)!
B
w+2 f
w (X):
Thenthefa tthatthe oeÆ ientsofX 1
andX w+1
inQ(X)vanishimplies
thatQ(X)2W
w
. Hen e
W
w
W
w
hf
w
i. Sin eitis learthatW
w
hf
w i
W
w
,we ompletetheproof.
Observingf
w 2
W
w
,we have thefollowing :
Lemma 2.3. (1)
W +
w
=W +
w .
(2)
W
w
=W
w
hf
w i.
Note that
W
w
b
V
w
from Lemma2.2.
3. The mapping . It is easy to see that 1 X w
belongs to W +
w
W
w
W
w
b
V
w
. So we may onsider the various quotient spa es, e.g.,
W +
=h1 X w
i,
W
w
=h1 X w
i, b
V
w
=h1 X w
i.
The purpose of thisse tion is to de ne a map q :V
w
! b
V
w
=h1 X w
i
whoseimageisexa tlythespa e
W
w
=h1 X w
i. LetV 0
w
denote thespa eof
polynomialsof degree w+1,namely , V 0
w
=f P
w+1
n=0
n X
n
:
n 2Kg.
Definition 3.1. (1)Let u:V
w
!V
w
bede nedby
uP(X)=Pj(1 U)(X) forP =P(X)2V
w :
(2) Letb:V
w
!V 0
w
bede ned by
bP(X)= w
X
n=0
n
n+1 B
n+1
(X+1)
where we write P(X) = P
w
n=0
n X
n
2 V
w
, with B
n
(X) denoting nth
Bernoullipolynomial.
(3) Lets:V 0
w
! b
V
w
bede nedby
sP(X)=Pj(1 S)(X);
more expli itly ,
sP(X)= w+1
X
n=0
n X
n w+1
X
n=0 ( 1)
n
n X
w n
forP(X)= P
w+1
n=0
n X
n
2V 0
w .
(4) Let :V
w
! b
V
w
be de nedby =sbu:
(5) Letq : b
V
w
! b
V
w
=h1 X w
i betheproje tionmap.
We needtwo lemmasto prove themaintheorem.
Lemma 3.1.
bP(X) bP(X 1)=P(X):
Proof. Set P(X)= P
w
n=0
n X
n
. Then
bP(X) bP(X 1)= w
X
n=0
n
n+1 (B
n+1
(X+1) B
n+1 (X))
= w
X
n=0
n
n+1
(n+1)X n
=P(X):
Thisfollows from thepropertyof Bernoullipolynomials:
(3:1) B
n+1
(X+1) B
n+1
(X)=(n+1)X n
:
Lemma 3.2.
(1)= 1
w+1 (X
w+1
( 1) w+1
X 1
)+(terms of degrees from 0 to w):
1
Proof.
(1)=sbu(1)=sb(1 X w
)=s
B
1
(X+1)
0+1
B
w+1
(X+1)
w+1
=s
B
1
(X)+X 0
B
w+1 (X)
w+1 X
w
=s
X w+1
w+1
+(terms ofdegrees from 0to w)
= 1
w+1 (X
w+1
( 1) w+1
X 1
)
+(terms of degreesfrom 0 to w):
Now we arereadyto des ribe theimageof thehomomorphism
q :V
w
! b
V
w
=h1 X w
i:
Theorem 3.3.
Imq =
W
w
=h1 X w
i:
Proof. Firstlyweshowthein lusionImq
W
w
=h1 X w
i,byproving
thatIm
W
w
. ForP(X)2V
w
,settingP
1
(X) =uP(X);we have
P
1
(X)=Pj(1 U)(X)=P(X) P
X 1
X
X w
:
Next let P
2
(X)=bP
1
(X). Then we have
P
2
(X) P
2
(X 1)=P
1 (X)
byLemma3.1. Furthermore,let P
3
(X)=sP
2
(X). Then we have
P
3
(X)=P
2
j(1 S)(X)=P
2
(X) P
2
1
X
X w
:
By thede nition,P
3
(X)= P(X).
Now we laim thatP
3
j(1+U+U 2
)=0. In fa t,
P
3
j(1+U +U 2
)(X)
=
P
2
(X) P
2
1
X
X w
j(1+U +U 2
)
=P
2
(X) P
2
1
X
X w
+P
2
X 1
X
X w
P
2
1
X 1
X
X 1
X
w
X w
+P
2
1
X 1
(X 1) w
P
2
1
1
1
X 1
w
(X 1) w
=P
2
(X) P
2
(X 1)+
P
2
X 1
X
P
2
X 1
X 1
X w
+
P
2
1
X 1
P
2
1
X 1
1
(X 1) w
=P
1
(X)+P
1
X 1
X
X w
+P
1
1
X 1
(X 1) w
=P
1
j(1+U+U 2
)(X)=Pj(1 U)(1+U +U 2
)(X)
=Pj(1 U 3
)(X)=0:
Thisshowsthat
(3:2) P(X)2Im fQ2 b
V
w
:Qj(1+U+U 2
)=0g:
Now notingthat1 S 2
=0,we have
(3:3) Ims=fQj(1 S):Q2 b
V
w
g=fQ2 b
V
w
:Qj(1+S)=0g:
Thus
(3:4) Im =ImsbufQ2 b
V
w
:Qj(1+S)=0g:
From (3.2) and (3.4), we obtain
(3:5) Im fQ2 b
V
w
:Qj(1+S)=Qj(1+U+U 2
)=0g=
W
w :
Thisgivesthe in lusionthatweare after,namely ,
(3:6) Imq q(
W
w )=
W
w
=h1 X w
i:
Se ondly we laim that q(
W
w
)Imq . Sin e (1) 2Im
W
w , and
(1)62W
w
byLemma3.2, we know
W
w
=W
w
h (1)i
notingthat W
w
is a odimension one subspa e of
W
w
. Hen e it suÆ es to
show thein lusionq(W
w
)Imq .
Let Q be any element of W
w
. We now show q(Q) 2 Imq . Sin e
Q 2ker(1+S) = Im(1 S), there is Q
1 2V
w
su h that Q
1
j(1 S)= Q,
namely ,
Q(X)=Q
1
(X) Q
1
1
X
X w
:
LetQ
2
(X) be de nedby
(3:7) Q
2
(X)=Q
1
(X) Q
1
(X 1):
Note thatQ 2V .
Next we showthat Q
2
j(1+U +U 2
)=0. In fa t,
(3:8) Q
2
j(1+U+U 2
)(X)
=Q
1
(X)+Q
1
X 1
X
X w
+Q
1
1
X 1
(X 1) w
Q
1
(X 1) Q
1
X 1
X 1
X w
Q
1
1
X 1
1
(X 1) w
=
Q
1
(X) Q
1
1
X
X w
+
Q
1
X 1
X
X w
Q
1
(X 1)
+
Q
1
1
X 1
(X 1) w
Q
1
X
X 1
(X 1) w
:
We also have
(3:9) Qj(1+U +U 2
)(X)
=Q(X)+Q
X 1
X
X w
+Q
1
X 1
(X 1) w
=Q
1
(X) Q
1
1
X
X w
+
Q
1
X 1
X
Q
1
X
X 1
X 1
X
w
X w
+
Q
1
1
X 1
Q
1
(X 1)
1
X 1
w
(X 1) w
:
Noti e that the expressions on the right hand sides of (3.8) and (3.9) do
oin ide,sothat we have
(3:10) Q
2
j(1+U +U 2
)=Qj(1+U +U 2
):
In parti ular, sin ethe right handsideof (3.10) is zero bythe assumption
thatQ2W
w
,thismeansQ
2
j(1+U+U 2
)=0 aswe required.
Sin e ker(1+U +U 2
) = Im(1 U), it follows that Q
2
2 Im(1 U).
Hen e there isQ
3
(X)2V
w
su hthat Q
3
j(1 U)=Q
2 .
Finally , we show thatq (Q
3
)=q(Q). Bythe de nitionsof and Q
2 ,
(3:11) q (Q
3
)=qsbu(Q
3
)=qsb(Q
2 ):
Sin e Q
1
(X) Q
1
(X 1)=Q
2
(X) by (3.7), andbQ
2
(X) bQ
2
(X 1) =
Q
2
(X) by Lemma 3.1, bQ
2 Q
1
is a onstant, say . Cal ulate the right
handsideof(3.11) to obtain
qsb(Q
2
)=qs(Q
1
+ )=q(sQ
1
+ (1 X w
))=q(Q+ (1 X w
))=q(Q)
asq((1 X w
))=0. Thismeansq (Q
3
)=q(Q).
Thus we have proved that, for any Q(X) 2 W
w
, there is Q
3
(X) 2 V
w
su h that q (Q
3
) = q(Q). This implies q(W
w
) Imq , ompleting the
4. Cal ulation. We al ulate ((X 1) n
) for n = 0;:::;w. Let ne
denote w nforn=0;:::;w. Firstnote that
u((X 1) n
)=(X 1) n
j(1 U)=(X 1) n
1
X
n
X w
=(X 1) n
( 1) n
X e n
;
moreover,
b(X n
)= 1
n+1 B
n+1
(X+1)
bythede nitionofb,and
b((X 1) n
)= 1
n+1 B
n+1 (X)
byLemma3.1and (3.1). Thuswehave
bu((X 1) n
)= B
n+1 (X)
n+1
( 1) n
B
e n+1
(X+1)
e n+1
:
HereweadoptKohnen{Zagier'snotationB 0
n
(X)forthenthBernoullipoly-
nomial withoutits B
1
-term ([5, p.208℄):
B 0
n (X)=
n
X
i=0
i6=1
n
i
B
i X
n i
= X
0in
ieven
n
i
B
i X
n i
:
Then we have
((X 1) n
)=sbu((X 1) n
)
= B
n+1 (X)
n+1 B
n+1
( 1 =X)
n+1 X
w
( 1) n
B
e n+1
(X+1)
e n+1
B
e n+1
( 1 =X+1)
e n+1
X w
= 1
n+1
B
n+1
(X) B
n+1
1
X
X w
( 1) n
e n+1
B
e n+1
(X)+(en+1)X e n
B
e n+1
1
X
X w
(en+1)
1
X
e n
X w
= 1
n+1
B 0
n+1
(X) B 0
n+1
1
X
X w
( 1) n
B 0
e n+1
(X) B 0
e n+1
1
X w
1
n+1
X n
+ 1
n+1
n+1
2
1
X X
w
+ ( 1)
n
e n+1
e n+1
2 X
e n
( 1) n
e n+1
e n+1
2
1
X
e n
X w
( 1) n
X e n
+( 1) n
1
X
e n
X w
= 1
n+1
B 0
n+1
(X) B 0
n+1
1
X
X w
( 1) n
e n+1
B 0
e n+1
(X) B 0
e n+1
1
X
X w
:
Summarizing theabove al ulation,we obtain
Lemma 4.1.
((X 1) n
)= 1
n+1
B 0
n+1
(X) B 0
n+1
1
X
X w
( 1) n
e n+1
B 0
e n+1
(X) B 0
e n+1
1
X
X w
:
5. The mapping . In Se tion 3, we de ned the mapping . In this
se tion,we willde ne and studya similarmapping :V
w
! b
V
w
. Firstlet
usde ne auxiliarymappingss 0
:V
w
!V
w , b
0
:V
w
!V 0
w and u
0
:V 0
w
! b
V
w
asfollows:
s 0
P(X)=Pj(1 S)(X);
b 0
P(X)= w
X
n=0
n
n+1 B
n+1
(X) forP(X)= w
X
n=0
n X
n
;
u 0
P(X)=Pj(1 U)(X)=P(X) P
X 1
X
X w
:
Note that, ifP(X)= P
w+1
n=0
n X
n
isan element of V 0
w ,then
P
X 1
X
X w
= w+1
X
n=0
n
(X 1) n
X w n
;
andithastermsofdegreerangingfrom 1tow+1. Thisimpliesu 0
P 2 b
V
w .
Nowlet usde nethemap :V
w
! b
V
w
byletting =u 0
b 0
s 0
:
Lemma 5.1.
(5:1) (X
n
)= ((X 1) n
) for n=0;:::;w:
Proof. We al ulate (X n
):
(X n
)=u 0
b 0
s 0
(X n
)=u 0
b 0
X n
1
n
X w
=u 0
b 0
(X n
( 1) n
X e n
)
=u 0
B
n+1 (X)
n+1
( 1) n
B
e n+1
(X)
e n+1
= 1
n+1
B
n+1
(X) B
n+1
X 1
X
X w
( 1) n
e n+1
B
e n+1
(X) B
e n+1
X 1
X
X w
= 1
n+1
B
n+1
(X) B
n+1
1
X
X w
(n+1)
1
X
n
X w
( 1) n
e n+1
B
e n+1
(X) B
e n+1
1
X
X w
(en+1)
1
X
e n
X w
= 1
n+1
B 0
n+1
(X) B 0
n+1
1
X
X w
n+1
2 X
n
+ n+1
2
1
X
n
X w
(n+1)( 1) n
X e n
( 1) n
e n+1
B 0
e n+1
(X) B 0
e n+1
1
X
X w
e n+1
2 X
e n
+ e n+1
2
1
X
e n
X w
(en+1)( 1) e n
X n
= 1
n+1
B 0
n+1
(X) B 0
n+1
1
X
X w
( 1) n
e n+1
B 0
e n+1
(X) B 0
e n+1
1
X
X w
= ((X 1) n
):
As a orollary ofthe above relation betweenthemappings and , we
andeterminetheimageofq whereq: b
V
w
! b
V
w
=h1 X w
iistheproje tion
mapas before.
Corollary5.2.
(5:2) Imq =Imq =
W
w
=h1 X w
i:
Proof. Let t : V
w
! V
w
be an isomorphism determined by t(X n
) =
(X 1) n
forn=0;:::;w. Then,byLemma5.1, wehave = t. Itfollows
thatImq =Imq astis anisomorphism.
Note that
B 0
n+1
1
X
= X
0in+1
n+1
i
B
i
1
X
n+1 i
=( 1) n+1
B 0
n+1
1
X
= B
n+1
(1=X); n odd,
B 0
n+1
(1=X); n even.
Then we obtainthe followingdes riptionfor (X n
) from Lemmas 4.1 and
5.1:
Lemma 5.3. (1) For n evenand 0nw,
(X n
)= 1
n+1 B
0
n+1 (X)+
X w
n+1 B
0
n+1
1
X
1
e n+1
B 0
e n+1
(X)
X w
e n+1
B 0
e n+1
1
X
:
(2) For n odd and 1nw 1,
(X n
)= 1
n+1 B
0
n+1 (X)
X w
n+1 B
0
n+1
1
X
+ 1
e n+1
B 0
e n+1
(X)
X w
e n+1
B 0
e n+1
1
X
:
6. Spanning sets of W
w and
W
w
.From Corollary5.2, we knowthat
bothfq (X n
)g w
n=0
and fq (X n
)g w
n=0 span
W
w
=h1 X w
i.
Sin e (X n
) is aneven(respe tively , odd)Laurent polynomialdepend-
ing on n being odd (respe tively , even), we an derive the following fa t
ratherplainly.
Lemma 6.1. (1) (X n
)2
W +
w
for nodd.
(2) (X n
)2
W
w
for neven.
In what follows, h (X n
)i
0nw;nodd(resp:even)
, denotes the spa e
spanned by (X n
) for n odd (resp. even) and 0 n w. (The notation
hr
(R
n )i
0nw;nodd(even)
will be used inthenext se tion denoting similar
spa es.) Forsubspa es V and W,V +W denotesthesubspa espannedby
V andW.
Lemma 6.2. (1) q(h (X n
)i
0nw;nodd )=q(
W +
w ).
(2) h (X n
)i
0nw;neven
=
W
w .
Proof. We know by Corollary 5.2 that Imq = q(
W
w ) = q(
W +
w )
q(
W
w
):Hen e,byLemma 6.1, we have
(6:1) q(h (X
n
)i
0nw;nodd )=q(
W +
w )
and
(6:2) q(h (X
n
)i
0nw;neven )=q(
W ):
Notethat qj
W
w :
W
w
!q(
W
w
) isan isomorphism. Thisisbe ause
W
w
=
W +
w
W
w
and h1 X w
i
W +
w
. Thuswe also have
(6:3) h (X
n
)i
0nw;neven
=
W
w
from (6.2).
ByLemma6.2,weobtainspanningsetsforW
w ,
W
w ,and
W +
w
=h1 X w
i.
Theorem 6.3. (1)W
w
(respe tively,
W
w
) isspanned by
(X n
)= 1
n+1 B
0
n+1 (X)+
X w
n+1 B
0
n+1
1
X
1
e n+1
B 0
e n+1
(X) X
w
e n+1
B 0
e n+1
1
X
for n evenand 2nw 2 (respe tively, 0nw).
(2) W +
w
=h1 X w
i=
W +
w
=h1 X w
i isspanned by
q (X n
)=q
1
n+1 B
0
n+1 (X)
X w
n+1 B
0
n+1
1
X
+ 1
e n+1
B 0
e n+1
(X) X
w
e n+1
B 0
e n+1
1
X
for n odd and 1nw 1.
Proof. By Lemma 6.2, the theorem is obvious ex ept for the ase of
W
w
. Sin e (X 0
) = (X w
) 62 W
w
, and (X n
) 2 W
w
for n even and
2nw 2,we know thatf (X n
)g
2nw 2;neven
spans W
w .
7.Relationsbetween (X n
)andr
(R
n
).Inthisse tionwewillshow
that (X n
)is relatedto thepolynomialr
(R
n
) studiedbyKohnen{Zagier
[5℄. Thisfa tleadsusto an alternativeproofof thetheoremof Ei hlerand
Shimuraon periodmappings.
Let us re all Kohnen{Zagier's polynomials r
(R
n
). Let S
w+2
denote
the spa e of usp forms of weight w+2 with respe t to SL
2
(Z). First let
r
n :S
w+2
!C bethemapping de nedby
r
n (f)=
1
\
0 f(it)t
n
dt;
Let r
(f) andr(f)be polynomials de nedby
r +
(f)(X)= X
0nw
neven ( 1)
n=2
w
n
r
n (f)X
w n
;
r (f)(X)= X
0nw
nodd ( 1)
(n 1)=2
w
n
r
n (f)X
w n
;
r(f)(X)= i1
\
0
f(z)(X z) w
dz
for f 2 S
w+2
. Then learly r(f) = ir +
(f)+r (f). Let R
n 2 S
w+2 be
de nedby
(f;R
n )=r
n
(f) foranyf 2S
w+2
where(;) denotesPetersson produ t.
ThefollowingisaresultofKohnen{Zagier[5℄whi hwasprovedapplying
Cohen's[2℄ representation ofR
n .
Theorem 7.1(Kohnen{Zagier). (1) For neven,0nw,
( 1)
(w+2)=2+n=2
2 w
r (R
n )(X)
= 1
n+1 B
0
n+1 (X)
X w
n+1 B
0
n+1
1
X
+ 1
e n+1
B 0
e n+1
(X)+ X
w
e n+1
B 0
e n+1
1
X
(Æ
e n;0
Æ
n;0 )
(w+2)!
(w+1)B
w+2 w+1
X
m= 1
modd B
m+1
(m+1)!
B
f m+1
(me +1)!
X m
:
(2) For n odd, 0nn,
( 1)
(w+2)=2+(n 1)=2
2 w
r +
(R
n )(X)
= 1
n+1 B
0
n+1 (X)
X w
n+1 B
0
n+1
1
X
+ 1
e n+1
B 0
e n+1
(X) X
w
e n+1
B 0
e n+1
1
X
w+2
B
w+2
B
n+1
n+1
B
e n+1
e n+1
(X w
1):
Comparing Theorem 7.1 and Lemma 5.3, we obtain relations between
(X n
) and r
(R ):
Proposition7.2. (1) For n even,0nw,
(X n
)= ( 1)
(w+2)=2+n=2
2 w
r (R
n )(X)
(Æ
e n ;0
Æ
n;0 )
(w+2)!
(w+1)B
w+2 f
w (X):
(2) For n odd, 0nw,
(X n
)=( 1)
(w+2)=2+(n 1)=2
2 w
r +
(R
n )(X)+
w+2
B
w+2
B
n+1
n+1
B
e n+1
e n+1
(X w
1):
Westudyrelationsbetweenthepolynomials (X n
)andr
(R
n
)further.
Lemma 7.3. (1) q(hr +
(R
n )i
0nw;nodd
)=q(h (X n
)i
0nw;nodd ) .
(2) hr (R
n )i
0nw;neven +hf
w
i=h (X n
)i
0nw;neven .
Proof. We rst show (1). Theequation in (2)of Proposition7.2 gives
riseto thefollowing ongruen e:
( 1)
(w+2)=2+(n 1)=2
2 w
r +
(R
n
)(X) (X n
)modh1 X w
i
fornodd. Thisimplies(1).
Next we show (2). Observing that f
w 2
W
w
and using Lemma 6.2(2),
we have
(7:1) h (X n
)i
0nw;neven +hf
w
i=h (X n
)i
0nw;neven :
It is learthat
(7:2) h (X n
)i
0nw;neven +hf
w
i=hr (R
n )i
0nw;neven +hf
w i
from Proposition7.2(1). From(7.1) and (7.2) weobtain
hr (R
n )i
0nw;neven +hf
w
i=h (X n
)i
0nw;neven
ompleting theproof of(2).
CombiningLemmas 6.2and 7.3weobtain:
Lemma 7.4. (1) q(
W +
w
)=q(hr +
(R
n )i
0nw;neven ).
(2)
W
w
=hr (R
n )i
0nw;neven +hf
w i .
We alsoobtainthe followinglemma:
Lemma 7.5. (1) q(W +
w
)=q(hr +
(R
n )i
0nw;neven ).
(2) W
w
=hr (R
n )i
0nw;neven .
Proof. Sin e W +
w
=
W +
w
, we have (1) from Lemma 7.4(1). From
Lemma7.4(2), we know that
W
w
isspannedbyfr (R
n )g
0nw;neven and
f
w
. Observingthatr (R
n
)arepolynomials,andthatW
w
isa odimension
one subspa eof
W ,we obtain(2).
Remark7.1. (a)In[5℄,thefa tthatfqr
(R n
)g w
n=0
isaspanningsetof
q(W
w
) (Lemma 7.5) is a onsequen eof the Ei hler{Shimuraisomorphism
forperiodmappings.
(b) Inourproofpresentedabove,wedo notneedto invoke thetheorem
of Ei hler and Shimura. As a matter of fa t, our Lemma 7.5 yields an
alternativeprooftothetheoremofEi hlerandShimuraonperiodmappings.
8. The theorem of Ei hler and Shimura
Corollary 8.1. r : S
w+2
! W
w
and qr +
: S
w+2
! W +
w
=h1 X w
i
are isomorphisms.
Proof. From Lemma7.5,weknowr andqr +
aresurje tive. It iswell
known thatthedimension ofS
w+2
is asfollows:
dimS
w+2
= 8
>
>
<
>
>
:
w+2
12
; w+2610 (mod12),
w+2
12
+1; w+210 (mod12).
Ontheother hand, asinLang[6℄,a linearalgebra argument shows
dimW
w
=dimW +
w
=h1 X w
i= 8
>
>
<
>
>
:
w+2
12
; w+2610 (mod12),
w+2
12
+1; w+210 (mod12).
Thisimpliesr and qr +
areisomorphisms.
Remark 8.1. Let M
w+2
denote the spa e of modular forms of weight
w+2. Zagier[7℄ \extended" the Ei hler{Shimura isomorphismto isomor-
phisms r +
:M
w+2
! W +
w
and qr : M
w+2
!
W
w
. As Lemma 7.5 gives
rise to the Ei hler{Shimuraisomorphism (Corollary 8.1), Lemma 7.4 gives
riseto Zagier'sisomorphisms.
Referen es
[1℄ Y.J.ChoieandD.Zagier,Rationalperiodfun tions,in:ATributetoEmilGross-
wald: NumberTheory and Related Analysis, M. Knopp and M. Sheingorn(eds.),
Contemp.Math.143,Amer.Math.So .,1993,89{108.
[2℄ H.Cohen,Sur ertainessommesdeserieslieesauxperiodesdeformesmodulaires,
in:Seminairedetheoriedenombres,Grenoble,1981.
[3℄ S.Fukuhara,Modularforms,generalizedDedekindsymbolsandperiodpolynomials,
preprint,1995.
[4℄ M. I.Knopp,Some new results onthe Ei hler ohomology of automorphi forms,
[5℄ W. Kohnen and D. Zagier, Modular forms with rational periods, in: Modular
Forms,R.A.Rankin(ed.),EllisHorwood,1984, 197{249.
[6℄ S.Lang,Introdu tion toModularForms,Springer,1976.
[7℄ D.Zagier,PeriodsofmodularformsandJa obithetafun tions,Invent.Math.104
(1991),449{465.
[8℄ |, Periodsof modular forms,tra es of He ke operators, and multiple zeta values,
preprint.
DepartmentofMathemati s
TsudaCollege
Tsuda-ma hi2-1-1
Kodaira-shi,Tokyo187,Japan
E-mail:fukuharatsuda.a .jp
Re eivedon6.11.1996
andinrevisedformon17.1.1997 (3071)
