Poisson Homology in Degree $0$ for some Rings of Symplectic Invariants

(1)

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Poisson Homology in Degree 0 for some Rings of

Symplectic Invariants

Frédéric Butin

To cite this version:

(2)

Poisson Homology in Degree

0

for some Rings of Symple ti Invariants Frédéri BUTIN

1

Abstra t

Let

g

be anite-dimensionalsemi-simpleLiealgebra,

h

aCartan subalgebra of

g

,and

W

itsWeyl group. The group

W

a tsdiagonallyon

V := h ⊕ h

∗

,aswellason

C[V ]

.Thepurposeofthisarti le istostudythePoisson homologyofthealgebraofinvariants

C[V ]

W

endowedwiththestandardsymple ti bra ket.

Tobegin with,wegivegeneralresults aboutthePoissonhomologyspa eindegree

0

,denotedby

HP

0 (C[V ]

W

₎

, inthe asewhere

g

isoftype

B

n

− C

n

or

D

n

,resultswhi hsupportAlev's onje ture.Thenwearefo usingthe interestontheparti ular asesofranks

2

and

3

,by omputingthePoissonhomologyspa eindegree

0

inthe ases where

g

isoftype

B

2

(

so

5

),

D

2

(

so

4

),then

B

3

(

so

7

),and

D

3 = A

3

(

so

6 ≃ sl

4

).Inordertodothis,wemakeuse ofafun tionalequationintrodu edbyY.Berest,P.EtingofandV.Ginzburg.Were over,byadierentmethod, the resultestablished by J.Alevand L. Foissy,a ording to whi hthe dimensionof

HP

0 (C[V ]

W

₎

equals

2

for

B

2

.Thenwe al ulatethedimensionofthisspa eandweshowthatitisequalto1for

D

2

.Wealso al ulateit fortherank

3

ases,weshowthatitisequalto

3

for

B

3 − C

3

and

1

for

D

3 = A

3

.

Key-words

Alev's onje ture;Pfa;Poissonhomology;Weylgroup;invariants;Berest-Etingof-Ginzburgequation.

1 Introdu tion

Let

G

beanitesubgroupofthesymple ti group

Sp(V )

,where

V

isa

C−

ve torspa eofdimension

2n

.Then the algebraof polynomialfun tions on

V

, denoted by

C[V ]

, is a Poisson algebra for the standard symple ti bra ket,and as

G

is asubgroupof thesymple ti group,the algebraofinvariants,denotedby

C[V ]

G

,isalso a Poissonalgebra.

Severalarti les weredevotedtothe omputation ofPoissonhomologyand ohomology ofthealgebraof invari-ants

C[V ]

G

. In parti ular, Y. Berest, P. Etingof and V. Ginzburg, in [BEG04℄, prove that the

0−

th spa e of Poissonhomologyof

C[V ]

G

isnite-dimensional.

Aftertheirworks[AL98 ℄and[AL98℄,J.Alev,M.A.Farinati,T.LambreandA.L.Solotarestablishafundamental resultin[AFLS00 ℄ :they omputeall thespa es ofHo hs hildhomology and ohomologyof

A

n

(C)

G

for every nitesubgroup

G

of

Sp

2n

C

.

Besides, J.Alevand L.Foissyshowin[AF06 ℄thatthe dimensionof thePoissonhomologyspa eindegree

0

of

C[h ⊕ h

∗

_]

W

isequal to theoneof the Ho hs hildhomology spa e indegree

0

of

A

2 (C)

W

, where

h

is aCartan subalgebra ofasemi-simpleLiealgebraofrank

2

withWeylgroup

W

.

Inthe following,givenanite-dimensionalsemi-simpleLiealgebra

g

,aCartansubalgebra

h

of

g

,and itsWeyl group

W

, we are interested in the Poisson homology of

C[V ]

G

in the ase where

V

is the symple ti spa e

V := h ⊕ h

∗

and

G := W

.

Thegroup

W

a ts diagonallyon

V

,this indu esana tion of

W

on

C[V ]

.Wedenote by

C[V ]

W

thealgebraof invariants underthis a tion.Endowedwith thestandard symple ti bra ket, thisalgebra is aPoisson algebra. Thea tionofthegroup

W

on

V

alsoindu esana tionof

W

ontheWeylalgebra

A

n

(C)

.

Tobeginwith,wewill givegeneralresultsaboutthePoissonhomologyspa eindegree

0

of

C[V ]

W

forthetypes

B

n

− C

n

and

D

n

,resultswhi hsupportAlev's onje tureandestablishaframeworkforapossibleproof.Se tions 2.3,2.4 and2.5 ontainthemainresultsofthisstudy.

Nextwewilluse theseresults inorderto ompletely al ulatethePoisson homologyspa e indegree

0

,denoted by

HP

0 (C[V ]

W

₎

,inthe asewhere

g

is

so

5

(i.e.

B

2

)sowere over,byadierentmethod,theresultestablished byJ.AlevandL.Foissyfor

so

5

inthearti le[AF06 ℄,namely

dim HP

0 (C[V ]

W

_{) = 2}

theninthe ase where

g

is

so

4

(i.e.

D

2 = A

1 × A

1

)byshowingthat

dim HP

0 (C[V ]

W

_{) = 1}

.Finallywewill provetheimportantproperty forrank

3

:

Proposition1 (Poissonhomologyindegree

0−

for

g

ofrank

3

) Let

HP

0 (C[V ]

W

₎

bethePoissonhomologyspa eindegree

0

of

C[V ]

W

and

HH

0 A

n

(C)

W

theHo hs hild homol-ogyspa e indegree

0

of

A

n

(C)

W

.

For

g

oftype

B

3

(

so

7

),we have

dim HP

0 C[V ]

W

_{= dim HH}

0 A

n

(C)

W

= 3

. For

g

oftype

D

3 = A

3

(

so

6 ≃ sl

4

),wehave

dim HP

0 C[V ]

W

_{= dim HH}

0 A

n

(C)

W

= 1

. 1

UniversitédeLyon,UniversitéLyon1,CNRS,UMR5208,InstitutCamilleJordan, 43blvddu11novembre1918,F-69622Villeurbanne-Cedex,Fran e

(3)

above.

2 Results about

B

n

− C

n

and

D

n

Asindi atedabove,weareinterestedinthePoissonhomologyof

C[h ⊕ h

∗

_]

W

,where

h

isaCartansubalgebraofa nite-dimensionalsemi-simpleLiealgebra

g

,and

W

itsWeylgroup.Wewillstudythetypes

B

n

and

D

n

.Re all thattherootsystemoftype

C

n

isdualtotherootsystemoftype

B

n

.SotheirWeylgroupsareisomorphi ,and thestudyofthe ase

C

n

isredu edtothestudyofthe ase

B

n

.

2.1 Denitions and notations

•

Set

S := C[x, y] = C[x

1 , . . . , x

n

, y

1 , . . . , y

n

]

.

For

m ∈ N

,wedenoteby

S(m)

theelementsof

S

ofdegree

m

. For

B

n

,wehave

W = (±1)

n

_{⋊ S}

n

= (±1)

n

· S

n

(permutationsofthevariablesandsign hangesofthevariables). For

D

n

,wehave

W = (±1)

n−1

_{⋊ S}

n

= (±1)

n−1

· S

n

(permutationsofthevariablesandsign hangesofaneven numberofvariables).

Everyelement

(a

1 , . . . , a

n

) ∈ (±1)

n

isidentiedwiththediagonalmatrix

Diag(a

1 , . . . , a

n

)

,andeveryelement

σ ∈ S

n

isidentiedwiththematrix

(δ

i,σ(j)

)

(i,j)∈[[1, n]]

.Wewilldenoteby

s

j

the

j−

thsign hange,i.e.

s

j

(x

k

) = x

k

if

k 6= j

,

s

j

(x

j

) = −x

j

,and

s

j

(y

k

) = y

k

if

k 6= j

,

s

j

(y

j

) = −y

j

. Asfortheelementsof

(±1)

n−1

,theyareidentiedwiththematri esoftheform

Diag((−1)

i

1 _{, (−1)}

i

1 +i

2 _{, (−1)}

i

2 +i

3 _{, . . . , (−1)}

i

_n−2

+i

_n−1

_{, (−1)}

i

_n−1

_),

with

i

k

∈ {0, 1}

.Wewilldenoteby

s

i,j

thesign hangeofthevariablesofindi es

i

and

j

. So,all theseelementsare in

O

n

C

,andbyidentifying

g ∈ W

with

g

0

0 g

,weobtain

W ⊂ Sp

2n

C

.

•

The(right)a tionof

W

on

S

isdenedfor

P ∈ S

and

g ∈ W

by

g · P (x, y) := P

n

X

j=1

g

1j

x

j

, . . . ,

n

X

j=1

g

nj

x

j

,

n

X

j=1

g

1j

y

j

, . . . ,

n

X

j=1

g

nj

y

j

!

.

Sowehave

h · (g · P ) = (gh) · P

.

Intheparti ular asewhere

σ ∈ S

n

,wehave

σ · P (x, y) = P (x

σ

−1

(1)

, . . . , x

σ

−1

(n)

, y

σ

−1

(1)

, . . . , y

σ

−1

(n)

)

.

•

On

S

,wedenethePoissonbra ket

{P, Q} := h∇P, ∇Qi = ∇P · (J ∇Q) = ∇

x

P · ∇

y

Q − ∇

y

P · ∇

x

Q,

where

h·, ·i

isthestandardsymple ti produ t,asso iatedtothematrix

J =

0 I

n

−I

n

0

. Asthe group

W

is asubgroupof

Sp(h·, ·i) = Sp

2n

C

,thealgebraofinvariants

S

W

isaPoisson algebraforthe bra ketdenedabove.

•

WedenetheReynoldsoperatorasthelinearmap

R

n

from

S

to

S

determinedby

R

n

(P ) =

1 |W |

X

g∈W

g · P.

Weset

A = C[z, t] = C[z

1 , . . . , z

n

, t

1 , . . . , t

n

]

and

S

′

_{:= A[x, y]}

,andweextendthemap

R

n

asa

A−

linearmap from

S

′

to

S

′

. Remark 2

Inthe aseof

B

n

,every elementof

S

W

hasanevendegree.(It isfalsefor

D

n

). 2.2 Poisson homology

•

Let

A

bea Poissonalgebra. We denoteby

Ω

p

_(A)

the

A−

module ofKählerdierentials, i.e. theve torspa e spannedbytheelementsoftheform

F

0 dF

1 ∧ · · · ∧ dF

p

,wherethe

F

j

belongto

A

,and

d : Ω

p

_{(A) → Ω}

p+1

_(A)

(4)

We onsiderthe omplex

. . .

∂

5 /

/

Ω

4 (A)

∂

4 /

/

Ω

3 (A)

∂

3 /

/

Ω

2 (A)

∂

2 /

/

Ω

1 (A)

∂

1 /

/

Ω

0 (A)

withBrylinsky-Koszulboundaryoperator

∂

p

(See[B88 ℄):

∂

p

(F

0 dF

1 ∧ · · · ∧ dF

p

)

=

p

X

j=1

(−1)

j+1

{F

0 , F

j

} dF

1 ∧ · · · ∧ d

dF

j

∧ · · · ∧ dF

p

+

X

1≤i<j≤p

(−1)

i+j

F

0 d{F

i

, F

j

} ∧ dF

1 ∧ · · · ∧ d

dF

i

∧ · · · ∧ d

dF

j

∧ · · · ∧ dF

p

.

ThePoissonhomologyspa eindegree

p

isgivenbytheformula

HP

p

(A) =

Ker

∂

p

/

Im

∂

p+1

.

Inparti ular,wehave

HP

0 (A) = A / {A, A}.

•

Inthesequel,wewilldenote

HP

0 (C[V ]

W

₎

by

HP

0 (W )

and

HH

0 (C[V ]

W

₎

by

HH

0 (W )

.

2.3 Ve tors of highest weight

0

Theaimofthis se tionis toshowthat

S

W

₍₂₎

is isomorphi to

sl

2

and thattheve torswhi hdonotbelongto

{S

W

_{, S}

W

_}

,areamongtheve torsofhighestweight

0

ofthe

sl

2 −

module

S

W

,anobservationwhi hwillsimplify the al ulations.

Proposition3

For

B

n

(

n ≥ 2

)and

D

n

(

n ≥ 3

), thesubspa e

S

W

₍₂₎

is isomorphi to

sl

2

. More pre isely

S

W

_{(2) = hE, F, Hi}

with therelations

{H, E} = 2E, {H, F } = −2F

and

{E, F } = H

, where theelements

E, F, H

are expli itly givenby

E =

n

2 R

n

(x

2

1 ) =

1

2 x

· x

F = −

n

2 R

n

(y

2

1 ) = −

2

1 y

· y

H = −n R

n

(x

1 y

1 ) = −x · y.

For

D

2

,wehave

S

W

_{(2) = hE, F, Hi ⊕ hE}

′

_{, F}

′

_{, H}

′

_i

(dire tsumoftwoLiealgebrasisomorphi to

sl

2

). Sothespa es

S

and

S

W

are

sl

2 −

modules. Proof:

Wedemonstratethepropositionfor

B

n

,theproofbeinganalogousfor

D

n

.

•

As

x

2

1

isinvariantundersign hanges,wemaywrite

R

n

(x

2

1 ) =

n!

1 P

σ∈S

n

σ · x

2

1

.Moreover,wehavethepartition

S

n

=

`

n

_j=1

A

j

,where

A

j

:= {σ ∈ S

n

/ σ(1) = j}

has ardinality

(n − 1)!

. Thus

R

n

(x

2

1 ) =

(n−1)!

n!

P

n

j=1

x

2 j

.Wepro eedlikewisewith

R

n

(y

2

1 )

and

R

n

(x

1 y

1 )

.

•

Weobviouslyhave

S

W

_{(2) ⊃ hE, F, Hi}

. Moreover,

R(x

2 j

) = R(x

2

1 )

,

R(y

2 j

) = R(y

1

2 )

and

R(x

j

y

j

) = R(x

1 y

1 )

.Last, if

i 6= j

, then

x

i

x

j

,

y

i

y

j

and

x

i

y

j

are mappedtotheiroppositebythe

i−

thsign hange

s

i

,and

W = s

i

· hs

1 , . . . , s

i−1

, s

i+1

, . . . , s

n

i · S

n

⊔ hs

1 , . . . , s

i−1

, s

i+1

, . . . , s

n

i · S

n

, thus

R

n

(x

i

x

j

) = R

n

(y

i

y

j

) = R

n

(x

i

y

j

) = 0

.Hen e

S

W

_{(2) ⊂ hE, F, Hi}

.

•

Wehave

∇E =

x

0 , ∇F =

0 −y

and

∇H =

−y

−x

, so

{E, F } = H, {H, E} = 2E

and

{H, F } = −2F

.

Wedenoteby

S

sl

2

thesetofve torsofhighestweight

0

,i.e.thesetofelementsof

S

whi hareannihilatedbythe a tionof

sl

2

.

For

j ∈ N

,wedenoteby

S(j)

sl

2

theelementsof

S

sl

2

ofdegree

j

.

As the a tion of

W

and the a tion of

sl

2

ommute, we may write

(S

W

₎

sl

₂

= (S

sl

₂

)

W

= S

sl

W

₂

and likewise for

S

W

_(j)

sl

₂

. Proposition4

•

If

S

W

ontainsnoelementofdegree

1

,thentheve torsofhighestweight

0

ofdegree

0

donotbelongto

{S

W

_{, S}

W

_}

.

•

Let

W

be oftype

B

n

(

n ≥ 2

)or

D

n

(

n ≥ 3

).If

S

W

ontainsnoelement of degree

1, 3,

, thentheve tors of highestweight

0

ofdegree

4

donot belongto

{S

W

_{, S}

W

_}

(5)

•

ThePoissonbra ketbeinghomogeneousofdegree

−2

,wehave

S

W

_{(0) ∩ {S}

W

_{, S}

W

_{} = {S}

W

_{(2), S}

W

_{(0)} = {0}}

.

•

Similarly,

S

W

_{(4) ∩ {S}

W

_{, S}

W

_{} = {S}

W

_{(0), S}

W

_{(6)} + {S}

W

_{(2), S}

W

_{(4)} = {sl}

2 , S

W

(4)}

,a ording to Proposi-tion3.Withthede ompositionofthe

sl

2 −

modules,wehave

S

W

_{(4) =}

L

m∈N

V (m)

,with

{sl

2 , V (m)} = V (m)

if

m ∈ N

∗

and

{sl

2 , V (0)} = {0}

.Soif

a ∈ S

W

_{(4) ∩ {S}

W

_{, S}

W

_}

,then

a ∈

L

m∈N

∗

V (m)

.

Proposition5

Forevery monomial

M = x

i

_y

j

,wehave

{H, M }

=

(|i| − |j|)M = (

deg

x

(M ) −

deg

y

(M ))M

{E, M }

=

P

n

_k=1

j

k

x

i

₁

1 . . . x

i

_k−1

k−1

x

i

_k

+1

k

x

i

k+1

. . . x

i

n

y

1 j

1 . . . y

j

_k−1

k−1

y

j

_k

−1

k

y

j

k+1

. . . y

j

n

,

{F, M }

=

P

n

_k=1

i

k

x

i

1

1 . . . x

i

_k−1

k−1

x

i

k

−1

k

x

i

k+1

. . . x

i

n

y

j

1

1 . . . y

j

_k−1

k−1

y

j

k

+1

k

y

j

k+1

. . . y

j

n

.

Inparti ular,everyve torofhighestweight

0

isofevendegree. Proof:Thisresultsfromasimple al ulation.

Remark 6

Let

j ∈ N

and

P ∈ S

W

_(j)

. A ording to thede omposition of

sl

2 −

module

S

W

_(j)

inweight subspa es, we may write

P =

P

m

k=−m

P

k

with

{H, P

k

} = k P

k

. Sowe have

S

W

(j)

{S

W

_{, S}

W

_}∩S

W

_(j)

=

S

W

(j)

_sl2

{S

W

_{, S}

W

_}∩S

W

_(j)

sl2

. Thustheve tors whi hdonot belongto

{S

W

_{, S}

W

_}

are tobefound amongtheve torsof highestweight

0

. ThefollowingpropertyisageneralizationofProposition3provedbyJ.AlevandL.Foissyin[AF06 ℄.Itenables ustoknowthePoin aréseriesofthealgebra

S

sl

2

.

Proposition7 For

l ∈ N

,we have

S

sl

2 (2l + 1) = {0}

and

dim S

sl

2 (2l) = (C

n−1

l+n−1

)

2 − C

n−1

l+n

C

n−1

l+n−2

.

ThefollowingresultisimportantforsolvingtheequationofBerest-Etingof-Ginzburg,be auseitgivesa des rip-tionof thespa eof ve torsof highestweight

0

,spa e inwhi hwewill sear hthesolutionsof thisequation.In theproofofthisproposition,weusethearti les[DCP76 ℄and[GK04 ℄ on erningthepfaanalgebras.

Proposition8

For

i 6= j

, set

X

i,j

:= x

i

y

j

− y

i

x

j

.Thenthealgebra

C[x, y]

sl

2

isthealgebra generated bythe

X

i,j

'sfor

(i, j) ∈

[[1, n]]

2

.Wedenotethisalgebraby

ChX

i,j

i

.

Thisalgebraisnotapolynomialalgebrafor

n ≥ 4

(e.g.

X

1,2

X

3,4

− X

1,3

X

2,4

+ X

2,3

X

1,4

= 0

). Proof:

•

Thein lusion

ChX

i,j

i ⊂ C[x, y]

sl

2

being obvious,allthatwehavetodoistoshowthatthePoin aréseriesof bothspa esareequal,knowingthattheonefor

C[x, y]

sl

2

isalreadygivenbyProposition7.

•

Consider the ve tors

u

j

:=

x

j

y

j

for

j = 1 . . . n

, inthe symple ti spa e

C

2

endowedwith the standard symple ti form

h·i

denedbythematrix

J :=

0

1 −1

0

.Let

{T

i,j

/ 1 ≤ i < j ≤ n}

beasetofindeterminates, andlet

T

e

betheantisymmetri matrixthegeneraltermofwhi his

T

i,j

if

i < j

.Then,a ording tose tion6of [DCP76 ℄,theideal

I

2

ofrelationsbetweenthe

hu

i

, u

j

i

(i.e.betweenthe

X

i,j

)isgeneratedbythepfaanminors of

T

e

ofsize

4 × 4

.Set

P F := ChX

i,j

i

.Sowehave

P F ≃ C[(T

i,j

)

i<j

] / I

2 =: P F

0

,i.e.thealgebra

P F

isisomorphi tothepfaanalgebra

P F

0

.ItsPoin aréseriesisgiveninse tion 4of[GK04 ℄by

dim P F

0 (m) = (C

m+n−2

m

)

2 − C

m+n−2

m−1

C

m+n−2

m+1

.

Sowehave

dim P F (2l) = (C

l

l+n−2

)

2 − C

l−1

l+n−2

C

l+1

l+n−2

.Weverifythat

dim P F (2l) = (C

_l+n−1

n−1

)

2 − C

_l+n

n−1

C

_l+n−2

n−1

= dim C[x, y]

sl

₂

(2l).

Wehaveobviously

dim P F (2l + 1) = 0 = dim C[x, y]

sl

(6)

Westudythefun tionnalequationintrodu edbyY.Berest,P.EtingofandV.Ginzburgin[BEG04℄.Thepoint isthatsolvingthisequation,inthespa e

S

W

sl

₂

,isequivalenttothedeterminationofthequotient

S

W

{S

W

, S

W

_}

,that istosaythe omputationofthePoissonhomologyspa eindegree

0

of

S

W

. Lemma 9 (Berest-Etingof-Ginzburg)

We onsider

C

2n

,endowedwithitsstandardsymple ti form,denoted by

h·, ·i

.Let

j ∈ N

. Let

S := C[x, y] = C[z]

,andlet

L

j

:=

S

W

(j)

{S

W

, S

W

_}∩S

W

_(j)

∗

bethelineardualof

S

W

(j)

{S

W

, S

W

_}∩S

W

_(j)

. Then

L

j

isisomorphi totheve torspa eof polynomials

P ∈ C[w]

W

_(j)

satisfying thefollowingequation:

∀ w, w

′

∈ C

2n

,

X

g∈W

hw, gw

′

i P (w + gw

′

) = 0

(1) Proof:

•

For

w

= (u, v) ∈ C

2n

and

z

= (x, y) ∈ C

2n

,weset

L

w

(z) :=

P

g∈W

e

hw, gzi

. Sowehave

{L

w

(z), L

w

′

(z)} = ∇

x

L

w

(z) · ∇

y

L

w

′

(z) − ∇

y

L

w

(z) · ∇

x

L

w

′

(z)

Wededu etheformula

{L

w

(z), L

w

′

(z)} =

X

g∈W

hw, gw

′

iL

w+gw

′

(z).

(2)

•

Moreover,

L

w

(z)

isapowerseriesin

w

,the oe ientsofwhi hgenerate

S

W

:

L

w

(z) =

∞

X

p=0

|W |

p!

R

n

"

_X

n

i=1

y

i

u

i

− x

i

v

i

!

p

#

(3) The oe ientsoftheseriesaretheimagesby

R

n

oftheelementsofthe anoni albasisof

S

.

Remark:inthe aseof

B

n

,thereisnoinvariantofodddegree,sowehave

L

w

(z) =

P

g∈W

h

(hw, gzi)

.

⊲

Set

M

p

(z) = {z

i

_{/ |i| = p}}

and

M

p

(w) = {w

i

_{/ |i| = p}}

. Foramonomial

m = x

i

_y

j

_{∈ M}

p

(z)

,let

m = u

e

j

_v

i

. Similarly,foramonomial

m = u

i

v

j

∈ M

p

(w)

,let

m = x

j

y

i

.

So,for

m ∈ M

p

(z)

,wehave

m = m

e

,andfor

m ∈ M

p

(w)

,wehave

m = m

e

. Theseries

L

w

(z)

maythenbewrittenas

L

w

(z) = |W | +

∞

X

j=1

X

m

j

∈M

j

(w)

α

m

j

R

n

(m

j

)m

j

= |W | +

∞

X

j=1

X

m

j

∈M

j

(z)

α

_g

m

j

R

n

(m

j

)

m

f

j

= |W | +

∞

X

j=1

L

j

w

(z),

(4) with

α

m

j

∈ Q

∗

.

⊲

Now

P

n

i=1

y

i

u

i

− x

i

v

i

p

=

P

_|a|+|b|=p

(−1)

|b|

C

a,b

p

x

b

y

a

u

a

v

b

,

where

C

a,b

p

=

a

1 !...a

n

p!

!b

1 !...b

n

!

isthemultinomial oe ient,thereforea ordingtoformula (3),wehave

L

w

(z) = |W | +

∞

X

p=1

X

|a|+|b|=p

(−1)

|b|

|W |

p!

C

a,b

p

R

n

x

b

y

a

u

a

v

b

(5)

By olle tingtheformulae(4)and(5),weobtain

α

u

a

v

b

= (−1)

|b|

|W |

p!

C

a,b

p

.

(6)

•

Weidentify

L

j

withtheve torspa eoflinearformson

S

W

_(j)

whi hvanishon

{S

W

_{, S}

W

_{} ∩ S}

W

_(j)

. Denethemap

π : L

j

→

{P ∈ C[w]

W

(j) / ∀ w, w

′

∈ C

2n

,

X

g∈W

hw, gw

′

i P (w + gw

′

) = 0}

f

7→

π

f

:= f (L

j

w

).

(7)

Then

π

is welldened:indeed

L

j

w

isapolynomialin

z

ofdegree

j

with oe ientsin

C[w]

,and expli itly,we have

f (L

j

w

) =

X

m

j

∈M

j

(z)

(7)

⊲

If two monomials

m

j

, m

′

j

∈ M

j

(z)

belong to a same orbit under the a tion of

S

n

, then the oe ients

α

m

_g

j

f (R

n

(m

j

))

and

α

g

m

′

j

f (R

n

(m

′

j

))

of

m

j

and

m

′

j

arethesame,thus

f (L

j

w

)

isinvariantunder

W

.

⊲

Besides,

π

f

is solution of

E

n

(P ) = 0

: indeed, we may extend

f

as a linear map dened on

S

W

{S

W

_{, S}

W

_}

=

L

∞

i=0

S

W

_(i)

{S

W

, S

W

_}∩S

W

_(i)

,bysetting

f = 0

on

S

W

_(i)

{S

W

, S

W

_}∩S

W

_(i)

for

i 6= j

. Then,a ordingto(2),wehavetheequality

0 = f ({L

w

, L

w

′

}) =

P

_g∈W

hw, gw

′

if (L

w+gw

′

)

, hen e

P

g∈W

hw, gw

′

_if

_L

j

w+gw

′

= 0

.So,thepolynomial

f (L

j

w

) ∈ C[w]

satisesequation(1).

•

Denethemap

ϕ : {P ∈ C[w]

W

_{(j) / ∀ w, w}

′

_{∈ C}

2n

_,

X

g∈W

hw, gw

′

i P (w + gw

′

) = 0}

→

L

j

P =

X

m

j

∈M

j

(w)

β

m

j

m

j

7→

ϕ

P

: R

n

(m

j

) 7→

β

_mj

α

_mj

.

(9)

⊲

For

f ∈ L

j

,wehave

ϕ

π

f

(R

n

(m

j

)) =

α

_mj

f

₍

R

n

(m

j

)

α

_mj

= f (R

n

(m

j

))

, thus

ϕ

π

f

= f

.

⊲

For

P =

X

m

j

∈M

j

(w)

β

m

j

m

j

∈ C[w]

W

(j)

,wehave

P =

X

m

j

∈M

j

(z)

β

_g

m

j

m

f

j

,soif

m

j

∈ M

j

(z)

,then

ϕ

P

(R

n

(m

j

)) =

β

_mj

_g

α

_mj

_g

. Consequently,

π

ϕ

_P

=

P

_m

j

∈M

j

(z)

α

g

m

j

ϕ

P

(R

n

(m

j

))

m

f

j

=

P

m

j

∈M

j

(z)

α

g

m

j

β

_mj

_g

α

_mj

_g

m

f

j

= P.

So

π

isbije tiveanditsinverseis

ϕ

.

⊲

Allwehavetodoistoshowthat

ϕ

P

vanisheson

{S

W

_{, S}

W

_{} ∩ S}

W

_(j)

. Let

P ∈ C[w]

W

_(j)

beasolutionofequation(1).Thenas,

π

ϕ

P

= P

,wehavefor

k + l = j

,

0 =

P

_g∈W

hw, gw

′

_{iP (w + gw}

′

_{) = ϕ}

P

{L

k

w

, L

l

w

′

}

But

{L

k

w

, L

l

w

′

} =

P

m

k

∈M

k

(w)

P

µ

l

∈M

l

(w

′

)

α

m

k

α

µ

l

m

k

µ

l

{R

n

(m

k

), R

n

(µ

l

)},

sothat

P

m

k

∈M

k

(w)

P

µ

l

∈M

l

(w

′

)

α

m

k

α

µ

l

m

k

µ

l

ϕ

P

({R

n

(m

k

), R

n

(µ

l

)}) = 0.

Thislastequalityisequivalentto

∀ k + l = j, ϕ

P

({R

n

(m

k

), R

n

(µ

l

)}) = 0,

whi hshows that

ϕ

P

vanisheson

{S

W

_{, S}

W

_{} ∩ S}

W

_(j)

.

Thefollowing orollaryenablesustomaketheequationofBerest-Etingof-Ginzburgmoreexpli it. Corollary 10

Weintrodu e

2n

indeterminates,denotedby

z

1 , . . . , z

n

, t

1 , . . . , t

n

,andweextendtheReynoldsoperatorinamap from

C[x, y, z, t]

toitselfwhi his

C[z, t]−

linear.Thentheve torspa e

L

j

isisomorphi totheve torspa eof polynomials

P ∈ S

W

_(j)

satisfyingthefollowingequation :

R

n

_X

n

i=1

z

i

y

i

− t

i

x

i

P (z

1 + x

1 , . . . , z

n

+ x

n

, t

1 + y

1 , . . . , t

n

+ y

n

)

= 0

(10) i.e.

E

n

(P ) := R

n

(z · y − t · x) P (x + z, y + t)

= 0

(11) Proof: Wehave

hw, w

′

_{i = w · (Jw}

′

_{) =}

P

n

i=1

(w

i

w

′

n+i

− w

n+i

w

′

i

)

.Thenequation(10)isequivalentto

(8)

X

g∈W

n

X

i=1

(z

i

n

X

j=1

g

ij

y

j

− t

i

n

X

j=1

g

ij

x

j

) P

z

1 +

n

X

j=1

g

1j

x

j

, . . . , z

n

+

n

X

j=1

g

nj

x

j

, t

1 +

n

X

j=1

g

1j

y

j

, . . . , t

n

+

n

X

j=1

g

nj

y

j

(13) iszero. Thisisequivalentto

n

X

i=1

(z

i

g · y

i

− t

i

g · x

i

)

X

g∈W

g ·

P (z

1 + x

1 , . . . , z

n

+ x

n

, t

1 + y

1 , . . . , t

n

+ y

n

)

= 0,

(14) thatistosay

R

n

X

i=1

z

i

y

i

− t

i

x

i

P (z

1 + x

1 , . . . , z

n

+ x

n

, t

1 + y

1 , . . . , t

n

+ y

n

)

= 0,

(15)

where

R

n

istheReynoldsoperatorextendedina

C[z, t]−

linearmap.

Remark 11

•

Caseof

B

n

: fora monomial

M ∈ C[x, y]

,

⊲

either there existsasign hangewhi hsends

M

toitsopposite,andthen

R

n

(M ) = 0

⊲

or

M

isinvariantunder everysign hangeandthen

R

n

(M ) =

P

σ∈S

n

σ · M

.

If

Q = R

n

(P )

with

P ∈ C[x, y]

,thenwe mayalwaysassume thatea h monomialof

P

,inparti ular

P

itself,is invariantunderthesign hanges.

•

Caseof

D

n

:wehavethesameresult,by onsideringthistimethesign hangesofanevennumberofvariables.

The aimof Proposition12and its orollary is to redu edrasti ally the spa e inwhi hwe sear hthe solutions of equation(11) :indeed, insteadof sear hingthe solutions in

S

W

,we maylimit ourselvestothe spa e ofthe elementswhi hareannihilatedbythea tionof

sl

2

.

Proposition12 Let

P ∈ C[x, y]

W

.We onsidertheelement

E

n

(P )

denedbytheformula(11)asapolynomialinthe indetermi-nates

z, t

andwith oe ientsin

C[x, y]

.

Thenthe oe ientof

z

1 t

1

in

E

n

(P )

is

−1

n

{H, P }

,thatof

t

2

1

is

−1

n

{E, P }

andthatof

z

2

1

is

1 n

{F, P }

. Proof:

We arryouttheprooffor

B

n

.Themethodisthesamefor

D

n

.

•

Wedenote by

c

z

1 t

1 (P )

the oe ient of

z

1 t

1

in

E

n

(P )

. Sin ethe maps

P 7→ c

z

1 t

1 (P )

and

P 7→ {H, P }

are linear, allwehavetodoistoprovethepropertyfor

P

oftheform

P = R

n

(M )

,where

M = x

i

_y

j

isamonomial whi hwemayassumeinvariantunderthesign hangesthankstoremark11.

Thentheformula(11)maybewritten

|W | E

n

(M )

=

|W | R

n

(z · y − t · x)(x + z)

i

(y + t)

j

=

X

c∈(±1)

n

X

σ∈S

n

c ·

"

(z

1 y

σ

−1

(1)

+ · · · + z

n

y

σ

−1

(n)

)

n

Y

k=1

(z

k

+ x

σ

−1

(k)

)

i

k

_(t

k

+ y

σ

−1

(k)

)

j

k

#

−

X

c∈(±1)

n

X

σ∈S

n

c ·

"

(t

1 x

σ

−1

(1)

+ · · · + t

n

x

σ

−1

(n)

)

n

Y

k=1

(z

k

+ x

σ

−1

(k)

)

i

k

_(t

k

+ y

σ

−1

(k)

)

j

k

#

•

Sothe oe ientof

z

1 t

1

isgivenby

(9)

Sin e

P =

1 n!

P

σ∈S

n

Q

n

k=1

x

i

σ(k)

k

y

j

σ(k)

k

,wededu ethat

n! c

z

1 t

1 (P )

=

X

σ∈S

n

c

z

1 t

1 n

Y

k=1

x

i

σ(k)

k

y

j

σ(k)

k

!

=

X

σ∈S

n

(j

σ(1)

− i

σ(1)

) R

n

Y

k=1

x

i

σ(k)

k

y

j

σ(k)

k

!

=

X

σ∈S

n

(j

σ(1)

− i

σ(1)

) R

n

(M ) = (n − 1)!

n

X

k=1

(j

k

− i

k

) R

n

(M )

=

(n − 1)! (

deg

y

(M ) −

deg

x

(M )) R

n

(M ) = −(n − 1)! {H, P }.

•

Wepro eedasfor

z

1 t

1

,bydenotingby

c

t

2

1 (P )

the oe ientof

t

2

1

in

E

n

(P )

.Thenwehave

|W | c

t

2

1 (M )

=

−

X

c∈(±1)

n

X

σ∈S

n

c ·

h

x

σ

−1

(1)

n

Y

k=1

x

i

k

σ

−1

(k)

!

j

1 y

_σ

j

1 −1

−1

(1)

n

Y

k=2

y

j

k

σ

−1

(k)

!

_i

=

−|(±1)

n

| j

1 X

σ∈S

n

x

i

1 +1

σ

−1

(1)

y

j

1 −1

σ

−1

(1)

n

Y

k=2

x

i

k

σ

−1

(k)

y

j

k

σ

−1

(k)

!

=

−|W | j

1 R

n

x

i

₁

1 +1

y

₁

j

1 −1

n

Y

k=2

x

i

k

y

j

_k

k

!!

.

Thus

n! c

t

2

1 (P )

=

X

σ∈S

n

c

t

2

1 n

Y

k=1

x

i

σ(k)

k

y

j

σ(k)

k

!

= −

X

σ∈S

n

j

σ(1)

R

n

x

i

σ(1)

+1

1 y

j

σ(1)

−1

1 n

Y

k=2

x

i

σ(k)

k

y

j

σ(k)

k

!!

=

−

n

X

p=1

X

σ∈S

n

σ(1)=p

j

σ(1)

R

n

x

i

σ(1)

+1

1 y

j

σ(1)

−1

1 n

Y

k=2

x

i

σ(k)

k

y

j

σ(k)

k

!!

=

−(n − 1)!

n

X

p=1

j

p

R

n

x

i

1

1 . . . x

i

p

+1

p

. . . x

i

n

y

j

1

1 . . . y

j

p

−1

p

. . . y

j

n

.

But

n! {E, P }

=

X

σ∈S

n

{E, x

i

σ(1)

1 . . . x

i

σ(n)

n

y

j

σ(1)

1 . . . y

j

σ(n)

n

}

=

X

σ∈S

n

X

p=1

j

σ(p)

x

i

σ(1)

1 . . . x

i

σ(p)

+1

p

. . . x

i

σ(n)

n

y

j

σ(1)

1 . . . y

j

σ(p)

−1

p

. . . y

j

σ(n)

n

=

n

X

p=1

n

X

q=1

X

σ∈S

n

σ(p)=q

j

σ(p)

x

i

σ(1)

1 . . . x

i

σ(p)

+1

p

. . . x

i

σ(n)

n

y

j

σ(1)

1 . . . y

j

σ(p)

−1

p

. . . y

j

σ(n)

n

=

n

X

q=1

j

q

n

X

p=1

X

σ∈S

n

σ(p)=q

x

i

σ(1)

1 . . . x

i

_σ(p)

+1

p

. . . x

i

_σ(n)

n

y

j

_σ(1)

1 . . . y

j

_σ(p)

−1

p

. . . y

j

_σ(n)

n

=

n!

n

X

q=1

j

q

R

n

x

i

1

1 . . . x

i

q

+1

q

. . . x

i

n

y

j

1

1 . . . y

j

q

−1

q

. . . y

j

n

.

So

c

t

2

1 (P ) =

−1

n

{E, P }

.Similarlyweshowthat

c

z

2

1 (P ) =

1 n

{F, P }

.

Corollary 13

•

Let

P ∈ C[x, y]

W

.If

P

satisesequation(11),then

P

isannihilatedby

sl

2

,i.e.

P ∈ S

W

sl

₂

.

•

Therefore the ve tor spa e

L

j

is isomorphi to the ve tor spa e of the polynomials

P ∈ S

W

sl

₂

(j)

satisfying equation(11).

Thusthedeterminationof

S

W

{S

W

, S

W

_}

=

S

W

sl2

{S

W

_{, S}

W

_}∩S

W

sl2

isequivalenttotheresolution,in

S

W

sl

₂

,ofequation(11). Proof:

Let

P ∈ C[x, y]

W

satisfyingequation(11).Thenallthe oe ientsofthepolynomial

E

n

(P ) ∈ (C[x, y])[z, t]

are zero.Inparti ular,a ordingtoProposition12,wehave

{H, P } = {E, P } = {F, P } = 0

.Hen e

P ∈ S

W

sl

₂

. These ondpointresultsfromCorollary10andfromtherstpoint.

(10)

Wedenethe (intermediate)map

s

n

int

: C[x, y, z, t]

→

C[x, y, z, t]

P

7→

P (0 y z t

1 , 0),

andwe set

E

n

int

(P ) := s

n

int

(E

n

(P )).

(16) Similarly,wedenethemap

s

n

: C[x, y, z, t]

→

C[x, y, z, t]

P

7→

P (0 y

1 , 0 z t

1 , 0),

andwe set

E

n

′

(P ) := s

n

(E

n

(P )).

(17) Thislastequation isequation (11)afterthesubstitution

x

1 = · · · = x

n

= y

2 = · · · = y

n

= t

2 = · · · = t

n

= 0

.

Remark 15

If

P

satisesequation(11),itsatisesobviouslyequation (17).

Inthe asewhere

n

isanoddinteger,theve torsofhighestweight

0

ofevendegreearethesamefor

B

n

and

D

n

, and equations (17) are identi al for bothtypes.Moreover, the linkbetweenequations (11) for

B

n

and

D

n

en-ablesustoprovetheinequality

dim HP

0 (D

n

) ≤ dim HP

0 (B

n

)

.Itisthepurposeofthetwofollowingpropositions. Proposition16

Byabuseofnotation,wedenoteby

S

B

n

_(2p)

(resp.

S

D

n

_(2p)

)thesetofinvariantelementsofdegree

2p

intype

B

n

(resp.

D

n

).Then we have

S

B

2n+1

_{(2p) = S}

D

2n+1

_(2p)

,

S

B

2n+1

sl

₂

(p) = S

D

2n+1

sl

₂

(p)

,and equations (17) in

S

B

2n+1

sl

₂

=

S

D

2n+1

sl

₂

asso iatedtobothtypesare thesame.

Thisresult isfalsefortheevenindi es: ounter-example :

dim S

D

4 sl

₂

(6) = 1

whereas

S

B

4 sl

₂

(6) = {0}

. Proof:

•

Weset

Φ

2n+1

(P ) =

X

σ∈S

2n+1

σ · P,

Ψ

B

2n+1

(P ) =

X

g∈(±1)

2n+1

g · P,

Ψ

D

2n+1

(P ) =

X

g∈(±1)

2n

g · P,

sothat

R

B

2n+1

(P ) =

1 |B

2n+1

|

Φ

2n+1

◦ Ψ

B

2n+1

,

and

R

D

2n+1

(P ) =

1 |D

2n+1

|

Φ

2n+1

◦ Ψ

D

2n+1

.

Weobviouslyhave

Ψ

B

2n+1

(S(2p)) ⊂ Ψ

D

2n+1

(S(2p))

. Conversely,sin e

Ψ

D

2n+1

(S(2p))

is spanned by the elementsof the form

Ψ

D

2n+1

(m)

with

m

∈ S(2p)

monomial, all we havetodoistoshowthat

Ψ

D

2n+1

(m)

belongsto

Ψ

B

2n+1

(S(2p))

,i.e.

Ψ

D

2n+1

(m)

is invariantunderthesign hanges. Now

m

= x

i

1

1 . . . x

i

2n+1

y

j

1

1 . . . y

j

2n+1

with

P

2n+1

k=1

(i

k

+ j

k

) = 2p

,therefore atleastoneofthe

i

k

+ j

k

is even.Let'sdenoteby

l

the orrespondingindex.

So,forevery

k 6= l

,wehave

s

k

(m) = (−1)

i

_k

+j

k

_m

_{= s}

k,l

(m)

and

s

l

(m) = m

.But

Ψ

D

2n+1

(m) =





X

q

1 =0,1...q

2n+1

=0,1

(−1)

q

1 [(i

1 +j

1 )+(i

2 +j

2 )]+q

2 [(i

2 +j

2 )+(i

3 +j

3 )]+···+q

2n

[(i

2n

+j

2n

)+(i

2n+1

+j

2n+1

)]





|

{z

}

a

m

m,

therefore

s

k

Ψ

D

2n+1

(m)

=

a

m

s

k,l

(m) = s

k,l

(a

m

m) = s

k,l

Ψ

D

2n+1

(m)

si

k 6= l

a

m

si

k = l

= Ψ

D

2n+1

(m)

.

•

Sowehave

S

D

n

_{(2p) = Φ}

2n+1

Ψ

D

2n+1

(S(2p))

= Φ

2n+1

Ψ

B

2n+1

(S(2p))

= S

B

n

_(2p)

. Hen e

S

B

n

sl

₂

(2p) = S

D

n

sl

₂

(2p)

.Besides,a ordingtoProposition5,

S

B

n

sl

₂

(2p + 1) = S

D

n

sl

₂

(2p + 1) = {0}

.

•

For

P ∈ S

B

n

sl

₂

,equation(17)maybewritten

(11)

Let

P

beinvariantbysign hanges.If

P

issolution ofequation (11)for

D

n

,then

P

issolution ofequation(11) for

B

n

.

Inparti ular,wehave

dim HP

0 (D

2n+1

) ≤ dim HP

0 (B

2n+1

)

. Proof:

Let

SB

n

(resp.

SD

n

)bethegroupofsign hangesof

B

n

(resp.

D

n

).Wemaywrite

SB

n

= SD

n

⊔ SD

n

· s

1

. Let

P

beinvariantbysign hanges.Sowehave

P = R

B

n

(P ) = R

D

n

(P )

,andequation(11)for

B

n

(resp.

D

n

)may bewritten

E

B

n

(P ) = R

B

n

(Q)

(resp.

E

D

n

(P ) = R

D

n

(Q)

),with

Q = (z · y − t · x) P (x + z y + t)

. If

P

issolutionofequation(11)for

D

n

,thenwehave:

R

B

n

(Q)

=

X

h∈SB

n

X

σ∈S

n

(σh) · Q =

X

h∈SB

n

h ·

X

σ∈S

n

σ · Q

!

=

X

g∈SD

n

g ·

X

σ∈S

n

σ · Q

!

+

X

g∈SD

n

(gs

1 ) ·

X

σ∈S

n

σ · Q

!

=

X

g∈SD

n

g ·

X

σ∈S

n

σ · Q

!

+ s

1

· "

X

g∈SD

n

g ·

X

σ∈S

n

σ · Q

!#

=

R

D

n

(Q) + s

1 · R

n

D

(Q) = 0.

So,

P

issolutionofequation(11)for

B

n

.

Wededu ethe laimedinequality,knowingthat,a ordingtoProposition16,

S

B

2n+1

sl

₂

= S

D

2n+1

sl

₂

.

2.5 Constru tion of graphs atta hed to the invariant polynomials Letus re alltheequalityofProposition8:

S

W

sl

₂

= R

n

(ChX

i,j

i)

.Moreover,a ording toCorollary 13,the om-putationof

HP

0 (S

W

₎

anberedu estosolvingEquation(11)inthespa e

S

W

sl

₂

.

Inordertohaveshorterandmorevisualnotations,werepresentthepolynomialsofthisspa ebygraphs,bythe methodexplainedindenition21.

Butbefore, letusquote,for theparti ular asethat weare interestedin, thefundamentalresultestablishedby J.Alev,M.A.Farinati,T.LambreandA.L.Solotarin[AFLS00℄:

Theorem 18 (Alev-Farinati-Lambre-Solotar) For

k = 0 . . . 2n

,thedimensionof

HH

k

(A

n

(C)

W

₎

isthenumberof onjuga y lassesof

W

admittingtheeigenvalue

1

withthemultipli ity

k

.

Byspe ializingtothe asesof

B

n

and

D

n

,weobtain: Corollary 19 (Alev-Farinati-Lambre-Solotar)

•

Fortype

B

n

,thedimensionof

HH

0 (A

n

(C)

W

₎

isthenumberof partitions

π(n)

of theinteger

n

.

•

Fortype

D

n

, thedimensionof

HH

0 (A

n

(C)

W

₎

isthenumberofpartitions

eπ(n)

oftheinteger

n

havinganeven number ofparts.

The onje tureofJ.Alevmaybesetforthasfollows : Conje ture 20 (Alev)

•

Forthetype

B

n

,thedimensionof

HP

0 (S

W

₎

equalsthenumberofpartitions

π(n)

oftheinteger

n

.

•

Forthe type

D

n

, the dimensionof

HP

0 (S

W

₎

equals thenumber of partitions

eπ(n)

of theinteger

n

having an evennumberofparts.

Now,letusshowhowto onstru t

π(n)

solutionsofequation(11)forthe aseof

B

n

. Denition 21

For

i 6= j

, wenote

X

i,j

= x

i

y

j

− y

i

x

j

. Toea helementof theform

M :=

Q

n−1

i=1

Q

n

j=i+1

X

2a

i,j

,we asso iate the(non-oriented)graph

g

Γ

M

su h that

⊲

theset ofverti esof

g

Γ

M

istheset ofindi es

{k ∈ [[1, n]] / ∃ i ∈ [[1, n]] / a

i,k

6= 0

or

a

k,i

6= 0}

,

⊲

twoverti es

i, j

of

Γ

g

M

are onne ted bytheedge

i

a

_i,j

j

if

a

i,j

6= 0

.

•

If

σ ∈ S

n

,thenthegraph

Γ

]

σ·M

isobtainedbypermutingtheverti esof

g

Γ

M

.

So,byrepla ingea hvertexbythesymbol

•

,weobtainagraph

Γ

M

su hthatthemap

M 7→ Γ

M

is onstanton ev-eryorbitunderthea tionof

B

n

(resp.

D

n

).Sowemayasso iatethisgraphtotheelement

R

n

(12)

Toalinear ombination

P

p

k=1

α

k

M

k

,weasso iatethegraph

P

p

k=1

α

k

Γ

M

k

.

•

We may extend this denition to elements of the form

M :=

Q

n−1

i=1

Q

n

j=i+1

X

b

i,j

by denoting an edge by

• bi,j

2

•

if

b

i,j

isevenandby

• bi,j +1

2

•

if

b

i,j

isodd.Be areful!Wehavefor example

• • = 0

.

Example 22 Thepolynomial

R

4 (X

4 1,2

X

1,3

2 X

1,4

2 )

isrepresented bythegraph

•

.

•

Ifagraph ontainsonlyevenedges(i.e.oftheform

• a

i,j

•

),thenitisrepresentedin

B

n

andin

D

n

bythe sameelement.

Thisresultisnotvalidinthe aseofgraphswhi h ontainoddedges:forexample,theelement

•

iszeroin

B

4

,butdierentfromzeroin

D

4

. Remark 23

The graphs orresponding topolynomialsobtainedbyoperatingtheReynoldsoperator fordierentindi esonthe same elementsof thealgebrageneratedbythe

X

i,j

'sarethesame.

For example, the elements

R

3 (X

2 1,2

)

and

R

44 (X

2 1,2

)

are represented in this way by the same graph

•

. Propositions25and26showthatthis hasnoee tonthestudy ofequation (11)for

B

n

.

Proposition24 Forevery

n ∈ N

∗

,the numberoflineargraphswithoutloopsandwithoutisolatedverti esisequaltothenumber of partitionsof

n

. (Amultipleedgeisviewedasaloop).

Proof:immediate.Toea hpartition

p

of theform

n = 1p

1 + 2p

2 + 3p

3 + · · · + np

n

,weasso iate thegraph having

p

j

linear onne ted omponentswith

j

verti es.

Proposition25

•

Let

P ∈ C[x

1 , . . . , x

n

, y

1 , . . . , y

n

]

. If

R

n

(P ) 6= 0

then

R

n+1

(P ) 6= 0

.

•

Let

P

1 , . . . , P

m

∈ C[x

1 , . . . , x

n

, y

1 , . . . , y

n

]

.If

R

n

(P

1 ), . . . , R

n

(P

m

)

arelinearlyindependent,then

R

n+1

(P

1 ), . . . , R

n+1

(P

m

)

are linearlyindependent.

Proof:

•

We arryouttheprooffor

B

n

.Wepro eedlikewisefor

D

n

.

Let

P ∈ C[x

1 , . . . , x

n

, y

1 , . . . , y

n

]

su hthat

R

n

(P ) 6= 0

.A ordingtoremark11,wemayassumethattheterms of

P

areinvariant undersign hanges.

We onsider theset

T

n

ofthetermsof

R

n

(P )

that wepartitioninto orbitsunderthea tionof

S

n

:so wehave theequality

T

n

=

`

r

j=1

O

j

.Consequently,

R

n

(P )

maybewritten

R

n

(P ) =

r

X

j=1

α

j

R

n

(M

j

),

where

M

j

∈ O

j

.

Let

c

n+1

:= (1, . . . , n + 1) ∈ S

n+1

,and

s

n+1

the

(n + 1)−

th sign hange, so that

B

n+1

= hs

n+1

, c

n+1

i · B

n

. Againbytheinvarian eundersign hanges, wededu e

R

n+1

(P ) =

1 n + 1

r

X

j=1

α

j

n

X

k=0

c

k

n+1

· R

n

(M

j

)

!

|

{z

}

t

_j

.

Nowif

i 6= j

,then

t

i

and

t

j

belongtotwodistin torbitsunderthea tionof

S

n+1

,Thereforethe

t

j

'sarelinearly independent.So

R

n+1

(P ) 6= 0

.

•

Letbe

P

1 , . . . , P

m

∈ C[x

1 , . . . , x

n

, y

1 , . . . , y

n

]

su hthat

R

n

(P

1 ), . . . , R

n

(P

m

)

arelinearlyindependent.Let's onsider a zero linear ombination

P

m

j=1

λ

j

R

n+1

(P

j

) = 0

, i.e.

R

n+1

P

m

j=1

λ

j

P

j