# The purpose of this note is to enhan e our under- standing of the spa e of period polynomials

## Full text

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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.

(2)

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:

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We mayalso onsiderthespa e

b

V = n

n

X

i=m

i X

i

:m;n2Zsu h thatmn;

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;

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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 em2

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.

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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

(6)

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

(7)

=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 =ImsbufQ2 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 .

(8)

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

(9)

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

:

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

0in

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

(10)

+ 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

)

(11)

=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

0in+1



n+1

i



B

i



1

X



n+1 i

=( 1) n+1

B 0

n+1



1

X



(12)

= 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 0nw,

(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 1nw 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

0nw;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

0nw;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

0nw;nodd )=q(

W +

w ).

(2) h (X n

)i

0nw;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

0nw;nodd )=q(

W +

w )

and

(6:2) q(h (X

n

)i

0nw;neven )=q(

W ):

(13)

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

0nw;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 2nw 2 (respe tively, 0nw).

(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 1nw 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

2nw 2,we know thatf (X n

)g

2nw 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;

(14)

Let r



(f) andr(f)be polynomials de nedby

r +

(f)(X)= X

0nw

neven ( 1)

n=2



w

n



r

n (f)X

w n

;

r (f)(X)= X

0nw

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,0nw,

( 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



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, 0nn,

( 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 ):

(15)

Proposition7.2. (1) For n even,0nw,

(X n

)= ( 1)

(w+2)=2+n=2

2 w

r (R

n )(X)

e n ;0

Æ

n;0 )

(w+2)!

(w+1)B

w+2 f

w (X):

(2) For n odd, 0nw,

(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

0nw;nodd

)=q(h (X n

)i

0nw;nodd ) .

(2) hr (R

n )i

0nw;neven +hf

w

i=h (X n

)i

0nw;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

0nw;neven +hf

w

i=h (X n

)i

0nw;neven :

It is learthat

(7:2) h (X n

)i

0nw;neven +hf

w

i=hr (R

n )i

0nw;neven +hf

w i

from Proposition7.2(1). From(7.1) and (7.2) weobtain

hr (R

n )i

0nw;neven +hf

w

i=h (X n

)i

0nw;neven

ompleting theproof of(2).

CombiningLemmas 6.2and 7.3weobtain:

Lemma 7.4. (1) q(

W +

w

)=q(hr +

(R

n )i

0nw;neven ).

(2)

W

w

=hr (R

n )i

0nw;neven +hf

w i .

We alsoobtainthe followinglemma:

Lemma 7.5. (1) q(W +

w

)=q(hr +

(R

n )i

0nw;neven ).

(2) W

w

=hr (R

n )i

0nw;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

0nw;neven and

f

w

. Observingthatr (R

n

)arepolynomials,andthatW

w

isa odimension

one subspa eof

W ,we obtain(2).

(16)

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+2610 (mod12),



w+2

12



+1; w+210 (mod12).

Ontheother hand, asinLang[6℄,a linearalgebra argument shows

dimW

w

=dimW +

w

=h1 X w

i= 8

>

>

<

>

>

:



w+2

12



; w+2610 (mod12),



w+2

12



+1; w+210 (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 ertainessommesdeserieslieesauxperiodesdeformesmodulaires,

in:Seminairedetheoriedenombres,Grenoble,1981.

[3℄ S.Fukuhara,Modularforms,generalizedDedekindsymbolsandperiodpolynomials,

preprint,1995.

[4℄ M. I.Knopp,Some new results onthe Ei hler ohomology of automorphi forms,

(17)

[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:fukuharatsuda.a .jp

Re eivedon6.11.1996

andinrevisedformon17.1.1997 (3071)

Updating...

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