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HAL Id: hal-00325594

https://hal.archives-ouvertes.fr/hal-00325594

Preprint submitted on 29 Sep 2008

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Poisson Homology in Degree 0 for some Rings of Symplectic Invariants

Frédéric Butin

To cite this version:

Frédéric Butin. Poisson Homology in Degree 0 for some Rings of Symplectic Invariants. 2008. �hal-

00325594�

(2)

Poisson Homology in Degree

0

for some Rings of Sympleti Invariants

FrédériBUTIN 1

Abstrat

Let

g

be anite-dimensionalsemi-simpleLiealgebra,

h

aCartan subalgebra of

g

,and

W

itsWeyl group. The

group

W

atsdiagonallyon

V := h ⊕ h

,aswellason

C[V ]

.Thepurposeofthisartile istostudythePoisson

homologyofthealgebraofinvariants

C[V ] W

endowedwiththestandardsympletibraket.

Tobegin with,wegivegeneralresults aboutthePoissonhomologyspaeindegree

0

,denotedby

HP 0 (C[V ] W )

,

intheasewhere

g

isoftype

B n − C n

or

D n

,resultswhihsupportAlev'sonjeture.Thenwearefousingthe

interestonthepartiularasesofranks

2

and

3

,byomputingthePoissonhomologyspaeindegree

0

intheases

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

ofafuntionalequationintroduedbyY.Berest,P.EtingofandV.Ginzburg.Wereover,byadierentmethod,

the resultestablished by J.Alevand L. Foissy,aording to whihthe dimensionof

HP 0 (C[V ] W )

equals

2

for

B 2

.Thenwealulatethedimensionofthisspaeandweshowthatitisequalto1for

D 2

.Wealsoalulateit

fortherank

3

ases,weshowthatitisequalto

3

for

B 3 − C 3

and

1

for

D 3 = A 3

.

Key-words

Alev'sonjeture;Pfa;Poissonhomology;Weylgroup;invariants;Berest-Etingof-Ginzburgequation.

1 Introdution

Let

G

beanitesubgroupofthesympletigroup

Sp(V )

,where

V

isa

C−

vetorspaeofdimension

2n

.Then

the algebraof polynomialfuntions on

V

, denoted by

C[V ]

, is a Poisson algebra for the standard sympleti

braket,and as

G

is asubgroupof thesympletigroup,the algebraofinvariants,denotedby

C[V ] G

,isalso a

Poissonalgebra.

Severalartiles weredevotedtotheomputation ofPoissonhomologyandohomology ofthealgebraofinvari-

ants

C[V ] G

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

0−

th spae of

Poissonhomologyof

C[V ] G

isnite-dimensional.

Aftertheirworks[AL98 ℄and[AL98℄,J.Alev,M.A.Farinati,T.LambreandA.L.Solotarestablishafundamental

resultin[AFLS00 ℄ :theyomputeall thespaes ofHohshildhomology andohomologyof

A n (C) G

for every

nitesubgroup

G

of

Sp 2n C

.

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

0

of

C[h ⊕ h ] W

isequal to theoneof the Hohshildhomology spae 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 sympleti spae

V := h ⊕ h

and

G := W

.

Thegroup

W

ats diagonallyon

V

,this induesanation of

W

on

C[V ]

.Wedenote by

C[V ] W

thealgebraof

invariants underthis ation.Endowedwith thestandard sympletibraket, thisalgebra is aPoisson algebra.

Theationofthegroup

W

on

V

alsoinduesanationof

W

ontheWeylalgebra

A n (C)

.

Tobeginwith,wewill givegeneralresultsaboutthePoissonhomologyspaeindegree

0

of

C[V ] W

forthetypes

B n − C n

and

D n

,resultswhihsupportAlev'sonjetureandestablishaframeworkforapossibleproof.Setions

2.3,2.4 and2.5ontainthemainresultsofthisstudy.

Nextwewilluse theseresults inorderto ompletelyalulatethePoisson homologyspae indegree

0

,denoted

by

HP 0 (C[V ] W )

,intheasewhere

g

is

so 5

(i.e.

B 2

)sowereover,byadierentmethod,theresultestablished byJ.AlevandL.Foissyfor

so 5

intheartile[AF06 ℄,namely

dim HP 0 (C[V ] W ) = 2

theninthease 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 )

bethePoissonhomologyspaeindegree

0

of

C[V ] W

and

HH 0 A n (C) W

theHohshildhomol-

ogyspae 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,Frane

email:butinmath.univ-lyon1.fr

(3)

above.

2 Results about

B n − C n

and

D n

Asindiatedabove,weareinterestedinthePoissonhomologyof

C[h ⊕ h ] W

,where

h

isaCartansubalgebraofa

nite-dimensionalsemi-simpleLiealgebra

g

,and

W

itsWeylgroup.Wewillstudythetypes

B n

and

D n

.Reall

thattherootsystemoftype

C n

isdualtotherootsystemoftype

B n

.SotheirWeylgroupsareisomorphi,and

thestudyofthease

C n

isreduedtothestudyofthease

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

(permutationsofthevariablesandsignhangesofthevariables).

For

D n

,wehave

W = (±1) n−1 ⋊ S n = (±1) n−1 · S n

(permutationsofthevariablesandsignhangesofaneven 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−

thsignhange,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

,theyareidentiedwiththematriesoftheform

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

thesignhangeofthevariablesofindies

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)ationof

W

on

S

isdenedfor

P ∈ S

and

g ∈ W

by

g · P (x, y) := P X n j=1

g 1j x j , . . . , X n j=1

g nj x j , X n j=1

g 1j y j , . . . , X n j=1

g nj y j

! .

Sowehave

h · (g · P ) = (gh) · P

.

Inthepartiularasewhere

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

,wedenethePoissonbraket

{P, Q} := h∇P, ∇Qi = ∇P · (J ∇Q) = ∇ x P · ∇ y Q − ∇ y P · ∇ x Q,

where

h·, ·i

isthestandardsympletiprodut,assoiatedtothematrix

J =

0 I n

−I n 0

.

Asthe group

W

is asubgroupof

Sp(h·, ·i) = Sp 2n C

,thealgebraofinvariants

S W

isaPoisson algebraforthe

braketdenedabove.

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

Intheaseof

B n

,every elementof

S W

hasanevendegree.(It isfalsefor

D n

).

2.2 Poisson homology

Let

A

bea Poissonalgebra. We denoteby

p (A)

the

A−

module ofKählerdierentials, i.e. thevetorspae spannedbytheelementsoftheform

F 0 dF 1 ∧ · · · ∧ dF p

,wherethe

F j

belongto

A

,and

d : Ω p (A) → Ω p+1 (A)

is

(4)

Weonsidertheomplex

. . . 5 / / 4 (A)

4

/ / 3 (A)

3

/ / 2 (A)

2

/ / 1 (A)

1

/ / 0 (A)

withBrylinsky-Koszulboundaryoperator

∂ p

(See[B88 ℄):

∂ p (F 0 dF 1 ∧ · · · ∧ dF p ) = X p j=1

(−1) j+1 {F 0 , F j } dF 1 ∧ · · · ∧ dF d 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 .

ThePoissonhomologyspaeindegree

p

isgivenbytheformula

HP p (A) =

Ker

∂ p /

Im

∂ p+1 .

Inpartiular,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 Vetors of highest weight

0

Theaimofthis setionis toshowthat

S W (2)

is isomorphito

sl 2

and thatthevetorswhihdonotbelongto

{S W , S W }

,areamongthevetorsofhighestweight

0

ofthe

sl 2 −

module

S W

,anobservationwhihwillsimplify thealulations.

Proposition3

For

B n

(

n ≥ 2

)and

D n

(

n ≥ 3

), thesubspae

S W (2)

is isomorphito

sl 2

. More preisely

S W (2) = hE, F, Hi

with therelations

{H, E} = 2E, {H, F } = −2F

and

{E, F } = H

, where theelements

E, F, H

are expliitly

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, H i ⊕ hE , F , H i

(diretsumoftwoLiealgebrasisomorphito

sl 2

).

Sothespaes

S

and

S W

are

sl 2 −

modules.

Proof:

Wedemonstratethepropositionfor

B n

,theproofbeinganalogousfor

D n

.

As

x 2 1

isinvariantundersignhanges,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}

hasardinality

(n − 1)!

.

Thus

R n (x 2 1 ) = (n−1)! n! P n

j=1 x 2 j

.Weproeedlikewisewith

R n (y 1 2 )

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 j 2 ) = 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−

thsignhange

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

.Hene

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

thesetofvetorsofhighestweight

0

,i.e.thesetofelementsof

S

whihareannihilatedbythe ationof

sl 2

.

For

j ∈ N

,wedenoteby

S(j) sl 2

theelementsof

S sl 2

ofdegree

j

.

As the ation of

W

and the ation 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

,thenthevetorsofhighestweight

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,

, thenthevetors of

highestweight

0

ofdegree

4

donot belongto

{S W , S W }

.

(5)

ThePoissonbraketbeinghomogeneousofdegree

−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)}

,aording toProposi-

tion3.Withthedeompositionofthe

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 k +1 x i k+1 k+1 . . . x i n n y 1 j 1 . . . y k−1 j k−1 y j k k −1 y k+1 j k+1 . . . y j n n , {F, M } = P n

k=1 i k x i 1 1 . . . x i k−1 k−1 x i k k −1 x i k+1 k+1 . . . x i n n y j 1 1 . . . y j k−1 k−1 y k j k +1 y k+1 j k+1 . . . y j n n .

Inpartiular,everyvetorofhighestweight

0

isofevendegree.

Proof:Thisresultsfromasimplealulation.

Remark 6

Let

j ∈ N

and

P ∈ S W (j)

. Aording to thedeomposition of

sl 2 −

module

S W (j)

inweight subspaes, 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) sl 2 {S W , S W }∩S W (j) sl 2

.

Thusthevetors whihdonot belongto

{S W , S W }

are tobefound amongthevetorsof highestweight

0

.

ThefollowingpropertyisageneralizationofProposition3provedbyJ.AlevandL.Foissyin[AF06 ℄.Itenables

ustoknowthePoinaréseriesofthealgebra

S sl 2

.

Proposition7

For

l ∈ N

,we have

S sl 2 (2l + 1) = {0}

and

dim S sl 2 (2l) = (C l+n−1 n−1 ) 2 − C l+n n−1 C l+n−2 n−1

.

ThefollowingresultisimportantforsolvingtheequationofBerest-Etingof-Ginzburg,beauseitgivesadesrip-

tionof thespaeof vetorsof highestweight

0

,spae inwhihwewill searhthesolutionsof thisequation.In

theproofofthisproposition,weusetheartiles[DCP76 ℄and[GK04 ℄onerningthepfaanalgebras.

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:

Theinlusion

ChX i,j i ⊂ C[x, y] sl 2

being obvious,allthatwehavetodoistoshowthatthePoinaréseriesof

bothspaesareequal,knowingthattheonefor

C[x, y] sl 2

isalreadygivenbyProposition7.

Consider the vetors

u j :=

x j

y j

for

j = 1 . . . n

, inthe sympleti spae

C 2

endowedwith the standard

sympletiform

h·i

denedbythematrix

J :=

0 1

−1 0

.Let

{T i,j / 1 ≤ i < j ≤ n}

beasetofindeterminates, andlet

T e

betheantisymmetrimatrixthegeneraltermofwhihis

T i,j

if

i < j

.Then,aording tosetion6of

[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

.ItsPoinaréseriesisgiveninsetion 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+n−2 l+1

.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 2 (2l + 1)

.HenetheequalityofthePoinaréseries.

(6)

WestudythefuntionnalequationintroduedbyY.Berest,P.EtingofandV.Ginzburgin[BEG04℄.Thepoint

isthatsolvingthisequation,inthespae

S sl W 2

,isequivalenttothedeterminationofthequotient

S W

{S W , S W }

,that

istosaytheomputationofthePoissonhomologyspaeindegree

0

of

S W

.

Lemma 9 (Berest-Etingof-Ginzburg)

Weonsider

C 2n

,endowedwithitsstandardsympleti 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

isisomorphitothevetorspaeof 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)

Wededuetheformula

{L w (z), L w (z)} = X

g∈W

hw, gw iL w+gw (z).

(2)

Moreover,

L w (z)

isapowerseriesin

w

,theoeientsofwhihgenerate

S W

:

L w (z) =

X ∞ p=0

|W | p! R n

" n X

i=1

y i u i − x i v i

! p #

(3)

Theoeientsoftheseriesaretheimagesby

R n

oftheelementsoftheanonialbasisof

S

.

Remark:intheaseof

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 e = u 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 e = m

,andfor

m ∈ M p (w)

,wehave

m e = m

.

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 p a,b x b y a u a v b ,

where

C p a,b = a p!

1 !...a n !b 1 !...b n !

isthemultinomial oeient,thereforeaordingtoformula (3),wehave

L w (z) = |W | + X ∞ p=1

X

|a|+|b|=p

(−1) |b| |W | p! C p a,b R n

x b y a

u a v b

(5)

Byolletingtheformulae(4)and(5),weobtain

α u a v b = (−1) |b| |W |

p! C a,b p .

(6)

Weidentify

L j

withthevetorspaeoflinearformson

S W (j)

whihvanishon

{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

withoeientsin

C[w]

,and expliitly,we

have

f(L j w ) = X

m j ∈M j (z)

α g m j f(R n (m j )) m f j ∈ C[w].

(8)

(7)

If two monomials

m j , m j ∈ M j (z)

belong to a same orbit under the ation of

S n

, then the oeients

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 , S W W } = L ∞

i=0

S W (i)

{S W , S W }∩S W (i)

,bysetting

f = 0

on

{S W , S S W W (i) }∩S W (i)

for

i 6= j

.

Then,aordingto(2),wehavetheequality

0 = f ({L w , L w }) = P

g∈W hw, gw if (L w+gw )

,

hene

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→ α β mj

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 )) = α β mj g

g mj

.

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

isbijetiveanditsinverseis

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,

whihshows that

ϕ P

vanisheson

{S W , S W } ∩ S W (j)

.

ThefollowingorollaryenablesustomaketheequationofBerest-Etingof-Ginzburgmoreexpliit.

Corollary 10

Weintrodue

2n

indeterminates,denotedby

z 1 , . . . , z n , t 1 , . . . , t n

,andweextendtheReynoldsoperatorinamap

from

C[x, y, z, t]

toitselfwhihis

C[z, t]−

linear.Thenthevetorspae

L j

isisomorphitothevetorspaeof

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

∀ w, w ∈ C, P

g∈W

P n

i=1 (w i P n

j=1 g ij w n+j − w n+i P n

j=1 g ij w j ) P

w 1 + P n

j=1 g 1j w j , . . . , w n + P n

j=1 g nj w j , w n+1 + P n

j=1 g 1j w n+j , . . . , w 2n + P n

j=1 g nj w n+j

= 0

(12)

(8)

X

g∈W

X n i=1

(z i

X n j=1

g ij y j − t i

X n j=1

g ij x j ) P z 1 +

X n j=1

g 1j x j , . . . , z n + X n j=1

g nj x j , t 1 + X n j=1

g 1j y j , . . . , t n + X n j=1

g nj y j

(13)

iszero.

Thisisequivalentto

X n 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 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,

(15)

where

R n

istheReynoldsoperatorextendedina

C[z, t]−

linearmap.

Remark 11

Caseof

B n

: fora monomial

M ∈ C[x, y]

,

either there existsasignhangewhihsends

M

toitsopposite,andthen

R n (M ) = 0

or

M

isinvariantunder everysignhangeandthen

R n (M ) = P

σ∈ S n σ · M

.

If

Q = R n (P )

with

P ∈ C[x, y]

,thenwe mayalwaysassume thateah monomialof

P

,inpartiular

P

itself,is

invariantunderthesignhanges.

Caseof

D n

:wehavethesameresult,byonsideringthistimethesignhangesofanevennumberofvariables.

The aimof Proposition12and itsorollary is to reduedrastially the spae inwhihwe searhthe solutions

of equation(11) :indeed, insteadof searhingthe solutions in

S W

,we maylimit ourselvestothe spae ofthe

elementswhihareannihilatedbytheationof

sl 2

.

Proposition12

Let

P ∈ C[x, y] W

.Weonsidertheelement

E n (P )

denedbytheformula(11)asapolynomialintheindetermi-

nates

z, t

andwithoeientsin

C[x, y]

.

Thentheoeientof

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

n 1 {F, P }

.

Proof:

Wearryouttheprooffor

B n

.Themethodisthesamefor

D n

.

Wedenote by

c z 1 t 1 (P )

the oeient of

z 1 t 1

in

E n (P )

. Sinethe 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

whihwemayassumeinvariantunderthesignhangesthankstoremark11.

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) ) Y n 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) ) Y n k=1

(z k + x σ − 1 (k) ) i k (t k + y σ − 1 (k) ) j k

#

Sotheoeientof

z 1 t 1

isgivenby

|W | c z 1 t 1 (M ) = X

c∈(±1) n

X

σ∈ S n

c · h y σ − 1 (1)

Y n k=1

x i σ k −1 (k)

!

j 1 y σ j 1 −1 1 (1)

Y n k=2

y σ j k −1 (k)

!

−x σ −1 (1) i 1 x i σ 1 −1 1 (1)

Y n k=2

x i σ k 1 (k)

! n Y

k=1

y σ j k 1 (k)

! i

= |(±1) n | X

σ∈ S n

(j 1 − i 1 ) Y n k=1

x i σ k −1 (k) y σ j k −1 (k)

!

= |W | (j 1 − i 1 ) R n (M ).

(9)

Sine

P = n! 1 P

σ∈ S n

Q n

k=1 x i k σ(k) y j k σ(k)

,wededuethat

n! c z 1 t 1 (P ) = X

σ∈ S n

c z 1 t 1

Y n k=1

x i k σ(k) y k j σ(k)

!

= X

σ∈ S n

(j σ(1) − i σ(1) ) R n

Y n k=1

x i k σ(k) y k j σ(k)

!

= X

σ∈ S n

(j σ(1) − i σ(1) ) R n (M) = (n − 1)!

X n k=1

(j k − i k ) R n (M)

= (n − 1)! (

deg

y (M ) −

deg

x (M)) R n (M ) = −(n − 1)! {H, P }.

Weproeedasfor

z 1 t 1

,bydenotingby

c t 2

1 (P)

theoeientof

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)

Y n k=1

x i σ k −1 (k)

!

j 1 y σ j 1 −1 −1 (1)

Y n k=2

y σ j k −1 (k)

! i

= −|(±1) n | j 1

X

σ∈ S n

x i σ 1 +1 1 (1) y σ j 1 −1 1 (1)

Y n k=2

x i σ k 1 (k) y σ j k 1 (k)

!

= −|W | j 1 R n x i 1 1 +1 y 1 j 1 −1 Y n k=2

x i k k y j k k

!!

.

Thus

n! c t 2

1 (P ) = X

σ∈ S n

c t 2 1

Y n k=1

x i k σ(k) y k j σ(k)

!

= − X

σ∈ S n

j σ(1) R n x i 1 σ(1) +1 y 1 j σ(1) −1 Y n k=2

x i k σ(k) y j k σ(k)

!!

= −

X n p=1

X

σ∈ S n σ(1)=p

j σ(1) R n x i 1 σ(1) +1 y j 1 σ(1) −1 Y n k=2

x i k σ(k) y k j σ(k)

!!

= −(n − 1)!

X n p=1

j p R n

x i 1 1 . . . x i p p +1 . . . x i n n y j 1 1 . . . y j p p −1 . . . y j n n

.

But

n! {E, P } = X

σ∈ S n

{E, x i 1 σ(1) . . . x i n σ(n) y 1 j σ(1) . . . y j n σ(n) }

= X

σ∈ S n

X n p=1

j σ(p) x i 1 σ(1) . . . x i p σ(p) +1 . . . x i n σ(n) y 1 j σ(1) . . . y j p σ(p) −1 . . . y j n σ(n)

= X n p=1

X n q=1

X

σ∈ S n σ(p)=q

j σ(p) x i 1 σ(1) . . . x i p σ(p) +1 . . . x i n σ(n) y j 1 σ(1) . . . y p j σ(p) −1 . . . y n j σ(n)

= X n q=1

j q

X n p=1

X

σ∈ S n σ(p)=q

x i 1 σ(1) . . . x i p σ(p) +1 . . . x i n σ(n) y 1 j σ(1) . . . y p j σ(p) −1 . . . y n j σ(n)

= n!

X n q=1

j q R n

x i 1 1 . . . x i q q +1 . . . x i n n y 1 j 1 . . . y q j q −1 . . . y j n 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 sl W 2

.

Therefore the vetor spae

L j

is isomorphi to the vetor spae of the polynomials

P ∈ S sl W 2 (j)

satisfying

equation(11).

Thusthedeterminationof

S W {S W , S W }

= S

W sl 2 {S W , S W }∩S W sl

2

isequivalenttotheresolution,in

S sl W 2

,ofequation(11).

Proof:

Let

P ∈ C[x, y] W

satisfyingequation(11).Thenalltheoeientsofthepolynomial

E n (P ) ∈ (C[x, y])[z, t]

are

zero.Inpartiular,aordingtoProposition12,wehave

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

.Hene

P ∈ S sl W 2

.

TheseondpointresultsfromCorollary10andfromtherstpoint.

WeendthissetionbydeningtwovariantsoftheequationofBerest-Etingof-Ginzburg:thesearetehnialtools

whihenableustoeliminatesomevariablesandthustosolveequation(11)moreeasily.

(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).

Intheasewhere

n

isanoddinteger,thevetorsofhighestweight

0

ofevendegreearethesamefor

B n

and

D n

,

and equations (17) are idential 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 sl 2n+1 2 (p) = S sl D 2 2n+1 (p)

,and equations (17) in

S sl B 2n+1 2 = S sl D 2 2n+1

assoiatedtobothtypesare thesame.

Thisresult isfalsefortheevenindies:ounter-example :

dim S sl D 2 4 (6) = 1

whereas

S sl B 4 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,sine

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 2n+1 y 1 j 1 . . . y 2n+1 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

theorrespondingindex.

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 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)

.

Hene

S sl B n 2 (2p) = S sl D 2 n (2p)

.Besides,aordingtoProposition5,

S B sl n 2 (2p + 1) = S sl D 2 n (2p + 1) = {0}

.

For

P ∈ S B sl n 2

,equation(17)maybewritten

2n+1 X

i=1

z i

X

σ∈ S n σ(1)=i

y 1 P (z 1 , . . . , z 2n+1 , t 1 , 0, . . . , 0, y 1

|{z}

i

, 0, . . . , 0) − y 1 P (z 1 , . . . , z 2n+1 , t 1 , 0, . . . , 0, −y 1

|{z}

i

, 0, . . . , 0)

 = 0.

Itisequation(17)for

P ∈ S sl D 2 n

.

(11)

Let

P

beinvariantbysignhanges.If

P

issolution ofequation (11)for

D n

,then

P

issolution ofequation(11)

for

B n

.

Inpartiular,wehave

dim HP 0 (D 2n+1 ) ≤ dim HP 0 (B 2n+1 )

.

Proof:

Let

SB n

(resp.

SD n

)bethegroupofsignhangesof

B n

(resp.

D n

).Wemaywrite

SB n = SD n ⊔ SD n · s 1

.

Let

P

beinvariantbysignhanges.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 n D (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

.

Wededuethelaimedinequality,knowingthat,aordingtoProposition16,

S sl B 2n+1 2 = S sl D 2 2n+1

.

2.5 Constrution of graphs attahed to the invariant polynomials

Letus realltheequalityofProposition8:

S sl W 2 = R n (ChX i,j i)

.Moreover,aording toCorollary 13,theom-

putationof

HP 0 (S W )

anbereduestosolvingEquation(11)inthespae

S W sl 2

.

Inordertohaveshorterandmorevisualnotations,werepresentthepolynomialsofthisspaebygraphs,bythe

methodexplainedindenition21.

Butbefore, letusquote,for thepartiularasethat 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 )

isthenumberofonjugaylassesof

W

admittingtheeigenvalue

1

withthemultipliity

k

.

Byspeializingtotheasesof

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

e π(n)

oftheinteger

n

havinganeven

number ofparts.

TheonjetureofJ.Alevmaybesetforthasfollows :

Conjeture 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

e π(n)

of theinteger

n

having an

evennumberofparts.

Now,letusshowhowtoonstrut

π(n)

solutionsofequation(11)fortheaseof

B n

.

Denition 21

For

i 6= j

, wenote

X i,j = x i y j − y i x j

.

Toeahelementof theform

M := Q n−1 i=1

Q n

j=i+1 X i,j 2a i,j

,we assoiate the(non-oriented)graph

g Γ M

suh that

theset ofvertiesof

g Γ M

istheset ofindies

{k ∈ [[1, n]] / ∃ i ∈ [[1, n]] / a i,k 6= 0

or

a k,i 6= 0}

,

twoverties

i, j

of

Γ g M

areonneted bytheedge

i

a i,j

j

if

a i,j 6= 0

.

If

σ ∈ S n

,thenthegraph

Γ ] σ·M

isobtainedbypermutingthevertiesof

g Γ M

.

So,byreplaingeahvertexbythesymbol

,weobtainagraph

Γ M

suhthatthemap

M 7→ Γ M

isonstantonev-

eryorbitundertheationof

B n

(resp.

D n

).Sowemayassoiatethisgraphtotheelement

R n Q n−1 i=1

Q n

j=i+1 X i,j 2a i,j

.

Referências

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