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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�
Poisson Homology in Degree
0
for some Rings of Sympleti InvariantsFrédériBUTIN 1
Abstrat
Let
g
be anite-dimensionalsemi-simpleLiealgebra,h
aCartan subalgebra ofg
,andW
itsWeyl group. Thegroup
W
atsdiagonallyonV := h ⊕ h ∗
,aswellasonC[V ]
.Thepurposeofthisartile istostudythePoissonhomologyofthealgebraofinvariants
C[V ] W
endowedwiththestandardsympletibraket.Tobegin with,wegivegeneralresults aboutthePoissonhomologyspaeindegree
0
,denotedbyHP 0 (C[V ] W )
,intheasewhere
g
isoftypeB n − C n
orD n
,resultswhihsupportAlev'sonjeture.Thenwearefousingtheinterestonthepartiularasesofranks
2
and3
,byomputingthePoissonhomologyspaeindegree0
intheaseswhere
g
isoftypeB 2
(so 5
),D 2
(so 4
),thenB 3
(so 7
),andD 3 = A 3
(so 6 ≃ sl 4
).Inordertodothis,wemakeuseofafuntionalequationintroduedbyY.Berest,P.EtingofandV.Ginzburg.Wereover,byadierentmethod,
the resultestablished by J.Alevand L. Foissy,aording to whihthe dimensionof
HP 0 (C[V ] W )
equals2
forB 2
.Thenwealulatethedimensionofthisspaeandweshowthatitisequalto1forD 2
.Wealsoalulateitfortherank
3
ases,weshowthatitisequalto3
forB 3 − C 3
and1
forD 3 = A 3
.Key-words
Alev'sonjeture;Pfa;Poissonhomology;Weylgroup;invariants;Berest-Etingof-Ginzburgequation.
1 Introdution
Let
G
beanitesubgroupofthesympletigroupSp(V )
,whereV
isaC−
vetorspaeofdimension2n
.Thenthe algebraof polynomialfuntions on
V
, denoted byC[V ]
, is a Poisson algebra for the standard sympletibraket,and as
G
is asubgroupof thesympletigroup,the algebraofinvariants,denotedbyC[V ] G
,isalso aPoissonalgebra.
Severalartiles weredevotedtotheomputation ofPoissonhomologyandohomology ofthealgebraofinvari-
ants
C[V ] G
. In partiular, Y. Berest, P. Etingof and V. Ginzburg, in [BEG04℄, prove that the0−
th spae ofPoissonhomologyof
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 everynitesubgroup
G
ofSp 2n C
.Besides, J.Alevand L.Foissyshowin[AF06 ℄thatthe dimensionof thePoissonhomologyspaeindegree
0
ofC[h ⊕ h ∗ ] W
isequal to theoneof the Hohshildhomology spae indegree0
ofA 2 (C) W
, whereh
is aCartansubalgebra ofasemi-simpleLiealgebraofrank
2
withWeylgroupW
.Inthe following,givenanite-dimensionalsemi-simpleLiealgebra
g
,aCartansubalgebrah
ofg
,and itsWeylgroup
W
, we are interested in the Poisson homology ofC[V ] G
in the ase whereV
is the sympleti spaeV := h ⊕ h ∗
andG := W
.Thegroup
W
ats diagonallyonV
,this induesanation ofW
onC[V ]
.Wedenote byC[V ] W
thealgebraofinvariants underthis ation.Endowedwith thestandard sympletibraket, thisalgebra is aPoisson algebra.
Theationofthegroup
W
onV
alsoinduesanationofW
ontheWeylalgebraA n (C)
.Tobeginwith,wewill givegeneralresultsaboutthePoissonhomologyspaeindegree
0
ofC[V ] W
forthetypesB n − C n
andD n
,resultswhihsupportAlev'sonjetureandestablishaframeworkforapossibleproof.Setions2.3,2.4 and2.5ontainthemainresultsofthisstudy.
Nextwewilluse theseresults inorderto ompletelyalulatethePoisson homologyspae indegree
0
,denotedby
HP 0 (C[V ] W )
,intheasewhereg
isso 5
(i.e.B 2
)sowereover,byadierentmethod,theresultestablished byJ.AlevandL.Foissyforso 5
intheartile[AF06 ℄,namelydim HP 0 (C[V ] W ) = 2
theninthease whereg
is
so 4
(i.e.D 2 = A 1 × A 1
)byshowingthatdim HP 0 (C[V ] W ) = 1
.Finallywewill provetheimportantpropertyforrank
3
:Proposition1 (Poissonhomologyindegree
0−
forg
ofrank3
)Let
HP 0 (C[V ] W )
bethePoissonhomologyspaeindegree0
ofC[V ] W
andHH 0 A n (C) W
theHohshildhomol-
ogyspae indegree
0
ofA n (C) W
.For
g
oftypeB 3
(so 7
),we havedim HP 0 C[V ] W
= dim HH 0 A n (C) W
= 3
.For
g
oftypeD 3 = A 3
(so 6 ≃ sl 4
),wehavedim 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
above.
2 Results about
B n − C n
andD n
Asindiatedabove,weareinterestedinthePoissonhomologyof
C[h ⊕ h ∗ ] W
,whereh
isaCartansubalgebraofanite-dimensionalsemi-simpleLiealgebra
g
,andW
itsWeylgroup.WewillstudythetypesB n
andD n
.Reallthattherootsystemoftype
C n
isdualtotherootsystemoftypeB n
.SotheirWeylgroupsareisomorphi,andthestudyofthease
C n
isreduedtothestudyoftheaseB n
.2.1 Denitions and notations
•
SetS := C[x, y] = C[x 1 , . . . , x n , y 1 , . . . , y n ]
.For
m ∈ N
,wedenotebyS (m)
theelementsofS
ofdegreem
.For
B n
,wehaveW = (±1) n ⋊ S n = (±1) n ·S n
(permutationsofthevariablesandsignhangesofthevariables).For
D n
,wehaveW = (±1) n−1 ⋊ S n = (±1) n−1 · S n
(permutationsofthevariablesandsignhangesofaneven numberofvariables).Everyelement
(a 1 , . . . , a n ) ∈ (±1) n
isidentiedwiththediagonalmatrixDiag(a 1 , . . . , a n )
,andeveryelementσ ∈ S n
isidentiedwiththematrix(δ i,σ(j) ) (i,j)∈[[1, n]]
.Wewilldenotebys j
thej−
thsignhange,i.e.s j (x k ) = x k
ifk 6= j
,s j (x j ) = −x j
,ands j (y k ) = y k
ifk 6= j
,s j (y j ) = −y j
.Asfortheelementsof
(±1) n−1
,theyareidentiedwiththematriesoftheformDiag((−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}
.Wewilldenotebys i,j
thesignhangeofthevariablesofindiesi
andj
.So,all theseelementsare in
O n C
,andbyidentifyingg ∈ W
withg 0 0 g
,weobtain
W ⊂ Sp 2n C
.•
The(right)ationofW
onS
isdenedforP ∈ S
andg ∈ W
byg · 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
,wehaveσ · P (x, y) = P (x σ − 1 (1) , . . . , x σ − 1 (n) , y σ − 1 (1) , . . . , y σ − 1 (n) )
.•
OnS
,wedenethePoissonbraket{P, Q} := h∇P, ∇Qi = ∇P · (J ∇Q) = ∇ x P · ∇ y Q − ∇ y P · ∇ x Q,
where
h·, ·i
isthestandardsympletiprodut,assoiatedtothematrixJ =
0 I n
−I n 0
.
Asthe group
W
is asubgroupofSp(h·, ·i) = Sp 2n C
,thealgebraofinvariantsS W
isaPoisson algebraforthebraketdenedabove.
•
WedenetheReynoldsoperatorasthelinearmapR n
fromS
toS
determinedbyR n (P ) = 1
|W | X
g∈W
g · P.
Weset
A = C[z, t] = C[z 1 , . . . , z n , t 1 , . . . , t n ]
andS ′ := A[x, y]
,andweextendthemapR n
asaA−
linearmapfrom
S ′
toS ′
.Remark 2
Intheaseof
B n
,every elementofS W
hasanevendegree.(It isfalseforD n
).2.2 Poisson homology
•
LetA
bea Poissonalgebra. We denotebyΩ p (A)
theA−
module ofKählerdierentials, i.e. thevetorspae spannedbytheelementsoftheformF 0 dF 1 ∧ · · · ∧ dF p
,wheretheF j
belongtoA
,andd : Ω p (A) → Ω p+1 (A)
isWeonsidertheomplex
. . . ∂ 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 ) = 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
isgivenbytheformulaHP p (A) =
Ker∂ p /
Im∂ p+1 .
Inpartiular,wehave
HP 0 (A) = A / {A, A}.
•
Inthesequel,wewilldenoteHP 0 (C[V ] W )
byHP 0 (W )
andHH 0 (C[V ] W )
byHH 0 (W )
.2.3 Vetors of highest weight
0
Theaimofthis setionis toshowthat
S W (2)
is isomorphitosl 2
and thatthevetorswhihdonotbelongto{S W , S W }
,areamongthevetorsofhighestweight0
ofthesl 2 −
moduleS W
,anobservationwhihwillsimplify thealulations.Proposition3
For
B n
(n ≥ 2
)andD n
(n ≥ 3
), thesubspaeS W (2)
is isomorphitosl 2
. More preiselyS W (2) = hE, F, Hi
with therelations
{H, E} = 2E, {H, F } = −2F
and{E, F } = H
, where theelementsE, F, H
are expliitlygivenby
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
,wehaveS W (2) = hE, F, H i ⊕ hE ′ , F ′ , H ′ i
(diretsumoftwoLiealgebrasisomorphitosl 2
).Sothespaes
S
andS W
aresl 2 −
modules.Proof:
Wedemonstratethepropositionfor
B n
,theproofbeinganalogousforD n
.•
Asx 2 1
isinvariantundersignhanges,wemaywriteR n (x 2 1 ) = n! 1 P
σ∈ S n σ · x 2 1
.Moreover,wehavethepartitionS n = ` n
j=1 A j
,whereA 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
.WeproeedlikewisewithR n (y 1 2 )
andR n (x 1 y 1 )
.•
WeobviouslyhaveS W (2) ⊃ hE, F, Hi
.Moreover,
R(x 2 j ) = R(x 2 1 )
,R(y j 2 ) = R(y 1 2 )
andR(x j y j ) = R(x 1 y 1 )
.Last, ifi 6= j
, thenx i x j
,y i y j
andx i y j
aremappedtotheiroppositebythe
i−
thsignhanges i
,andW = 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
.HeneS 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
thesetofvetorsofhighestweight0
,i.e.thesetofelementsofS
whihareannihilatedbythe ationofsl 2
.For
j ∈ N
,wedenotebyS(j) sl 2
theelementsofS sl 2
ofdegreej
.As the ation of
W
and the ation ofsl 2
ommute, we may write(S W ) sl 2 = (S sl 2 ) W = S sl W 2
and likewisefor
S W (j) sl 2
.Proposition4
•
IfS W
ontainsnoelementofdegree1
,thenthevetorsofhighestweight0
ofdegree0
donotbelongto{S W , S W }
.•
LetW
be oftypeB n
(n ≥ 2
)orD n
(n ≥ 3
).IfS W
ontainsnoelement of degree1, 3,
, thenthevetors ofhighestweight
0
ofdegree4
donot belongto{S W , S W }
.•
ThePoissonbraketbeinghomogeneousofdegree−2
,wehaveS 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,wehaveS W (4) = L
m∈ N V (m)
,with{sl 2 , V (m)} = V (m)
ifm ∈ N ∗
and{sl 2 , V (0)} = {0}
.Soifa ∈ S W (4) ∩ {S W , S W }
,thena ∈ L
m∈ N ∗ V (m)
.Proposition5
Forevery monomial
M = x i y j
,wehave{H, M } = (|i| − |j|)M = (
degx (M ) −
degy (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
andP ∈ S W (j)
. Aording to thedeomposition ofsl 2 −
moduleS W (j)
inweight subspaes, we maywrite
P = P m
k=−m P k
with{H, P k } = k P k
. Sowe haveS 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 highestweight0
.ThefollowingpropertyisageneralizationofProposition3provedbyJ.AlevandL.Foissyin[AF06 ℄.Itenables
ustoknowthePoinaréseriesofthealgebra
S sl 2
.Proposition7
For
l ∈ N
,we haveS sl 2 (2l + 1) = {0}
anddim 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.Intheproofofthisproposition,weusetheartiles[DCP76 ℄and[GK04 ℄onerningthepfaanalgebras.
Proposition8
For
i 6= j
, setX i,j := x i y j − y i x j
.ThenthealgebraC[x, y] sl 2
isthealgebra generated bytheX i,j
'sfor(i, j) ∈ [[1, n]] 2
.WedenotethisalgebrabyChX 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:
•
TheinlusionChX i,j i ⊂ C[x, y] sl 2
being obvious,allthatwehavetodoistoshowthatthePoinaréseriesofbothspaesareequal,knowingthattheonefor
C[x, y] sl 2
isalreadygivenbyProposition7.•
Consider the vetorsu j :=
x j
y j
for
j = 1 . . . n
, inthe sympleti spaeC 2
endowedwith the standardsympletiform
h·i
denedbythematrixJ :=
0 1
−1 0
.Let
{T i,j / 1 ≤ i < j ≤ n}
beasetofindeterminates, andletT e
betheantisymmetrimatrixthegeneraltermofwhihisT i,j
ifi < j
.Then,aording tosetion6of[DCP76 ℄,theideal
I 2
ofrelationsbetweenthehu i , u j i
(i.e.betweentheX i,j
)isgeneratedbythepfaanminorsof
T e
ofsize4 × 4
.SetP F := ChX i,j i
.SowehaveP F ≃ C[(T i,j ) i<j ] / I 2 =: P F 0
,i.e.thealgebraP F
isisomorphitothepfaanalgebra
P F 0
.ItsPoinaréseriesisgiveninsetion 4of[GK04 ℄bydim 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
.Weverifythatdim 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.WestudythefuntionnalequationintroduedbyY.Berest,P.EtingofandV.Ginzburgin[BEG04℄.Thepoint
isthatsolvingthisequation,inthespae
S sl W 2
,isequivalenttothedeterminationofthequotientS W
{S W , S W }
,thatistosaytheomputationofthePoissonhomologyspaeindegree
0
ofS W
.Lemma 9 (Berest-Etingof-Ginzburg)
Weonsider
C 2n
,endowedwithitsstandardsympleti form,denoted byh·, ·i
.Letj ∈ N
.Let
S := C[x, y] = C[z]
,andletL 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 polynomialsP ∈ C[w] W (j)
satisfying thefollowingequation:∀ w, w ′ ∈ C 2n , X
g∈W
hw, gw ′ i P (w + gw ′ ) = 0
(1)Proof:
•
Forw = (u, v) ∈ C 2n
andz = (x, y) ∈ C 2n
,wesetL 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)
isapowerseriesinw
,theoeientsofwhihgenerateS 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
oftheelementsoftheanonialbasisofS
.Remark:intheaseof
B n
,thereisnoinvariantofodddegree,sowehaveL w (z) = P
g∈W
h(hw, gzi)
.⊲
SetM p (z) = {z i / |i| = p}
andM p (w) = {w i / |i| = p}
.Foramonomial
m = x i y j ∈ M p (z)
,letm e = u j v i
.Similarly,foramonomial
m = u i v j ∈ M p (w)
,letm = x j y i
.So,for
m ∈ M p (z)
,wehavem e = m
,andform ∈ M p (w)
,wehavem e = m
.Theseries
L w (z)
maythenbewrittenasL 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 ∗
.⊲
NowP 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 ,
whereC p a,b = a p!
1 !...a n !b 1 !...b n !
isthemultinomial oeient,thereforeaordingtoformula (3),wehaveL 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)•
WeidentifyL j
withthevetorspaeoflinearformsonS 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:indeedL j w
isapolynomialinz
ofdegreej
withoeientsinC[w]
,and expliitly,wehave
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)⊲
If two monomialsm j , m ′ j ∈ M j (z)
belong to a same orbit under the ation ofS n
, then the oeientsα m g j f(R n (m j ))
andα g
m ′ j f(R n (m ′ j ))
ofm j
andm ′ j
arethesame,thusf(L j w )
isinvariantunderW
.⊲
Besides,π f
is solution ofE n (P ) = 0
: indeed, we may extendf
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)
,bysettingf = 0
on{S W , S S W W (i) }∩S W (i)
fori 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,thepolynomialf(L j w ) ∈ C[w]
satisesequation(1).•
Denethemapϕ : {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)⊲
Forf ∈ L j
,wehaveϕ 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
.⊲
ForP = X
m j ∈M j (w)
β m j m j ∈ C[w] W (j)
,wehaveP = X
m j ∈M j (z)
β g m j m f j
,soifm j ∈ M j (z)
,thenϕ 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
isbijetiveanditsinverseisϕ
.⊲
Allwehavetodoistoshowthatϕ P
vanisheson{S W , S W } ∩ S W (j)
.Let
P ∈ C[w] W (j)
beasolutionofequation(1).Thenas,π ϕ P = P
,wehavefork + 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,denotedbyz 1 , . . . , z n , t 1 , . . . , t n
,andweextendtheReynoldsoperatorinamapfrom
C[x, y, z, t]
toitselfwhihisC[z, t]−
linear.ThenthevetorspaeL j
isisomorphitothevetorspaeofpolynomials
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)
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
istheReynoldsoperatorextendedinaC[z, t]−
linearmap.Remark 11
•
CaseofB n
: fora monomialM ∈ C[x, y]
,⊲
either there existsasignhangewhihsendsM
toitsopposite,andthenR n (M ) = 0
⊲
orM
isinvariantunder everysignhangeandthenR n (M ) = P
σ∈ S n σ · M
.If
Q = R n (P )
withP ∈ C[x, y]
,thenwe mayalwaysassume thateah monomialofP
,inpartiularP
itself,isinvariantunderthesignhanges.
•
CaseofD 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 oftheelementswhihareannihilatedbytheationof
sl 2
.Proposition12
Let
P ∈ C[x, y] W
.WeonsidertheelementE n (P )
denedbytheformula(11)asapolynomialintheindetermi-nates
z, t
andwithoeientsinC[x, y]
.Thentheoeientof
z 1 t 1
inE n (P )
is−1 n {H, P }
,thatoft 2 1
is−1 n {E, P }
andthatofz 2 1
isn 1 {F, P }
.Proof:
Wearryouttheprooffor
B n
.ThemethodisthesameforD n
.•
Wedenote byc z 1 t 1 (P )
the oeient ofz 1 t 1
inE n (P )
. Sinethe mapsP 7→ c z 1 t 1 (P )
andP 7→ {H, P }
arelinear, allwehavetodoistoprovethepropertyfor
P
oftheformP = R n (M)
,whereM = x i y j
isamonomialwhihwemayassumeinvariantunderthesignhangesthankstoremark11.
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
#
•
Sotheoeientofz 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 ).
Sine
P = n! 1 P
σ∈ S n
Q n
k=1 x i k σ(k) y j k σ(k)
,wededuethatn! 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)! (
degy (M ) −
degx (M)) R n (M ) = −(n − 1)! {H, P }.
•
Weproeedasforz 1 t 1
,bydenotingbyc t 2
1 (P)
theoeientoft 2 1
inE 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 }
.Similarlyweshowthatc z 2
1 (P ) = 1 n {F, P }
.Corollary 13
•
LetP ∈ C[x, y] W
.IfP
satisesequation(11),thenP
isannihilatedbysl 2
,i.e.P ∈ S sl W 2
.•
Therefore the vetor spaeL j
is isomorphi to the vetor spae of the polynomialsP ∈ S sl W 2 (j)
satisfyingequation(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).ThenalltheoeientsofthepolynomialE n (P ) ∈ (C[x, y])[z, t]
arezero.Inpartiular,aordingtoProposition12,wehave
{H, P } = {E, P } = {F, P } = 0
.HeneP ∈ S sl W 2
.TheseondpointresultsfromCorollary10andfromtherstpoint.
WeendthissetionbydeningtwovariantsoftheequationofBerest-Etingof-Ginzburg:thesearetehnialtools
whihenableustoeliminatesomevariablesandthustosolveequation(11)moreeasily.
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,thevetorsofhighestweight0
ofevendegreearethesameforB n
andD n
,and equations (17) are idential for bothtypes.Moreover, the linkbetweenequations (11) for
B n
andD 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)
)thesetofinvariantelementsofdegree2p
intypeB n
(resp.
D n
).Then we haveS 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) inS 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
whereasS sl B 4 2 (6) = {0}
.Proof:
•
WesetΦ 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 ,
andR 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 formΨ D 2n+1 (m)
withm ∈ S(2p)
monomial,all we havetodoistoshowthat
Ψ D 2n+1 (m)
belongstoΨ B 2n+1 (S(2p))
,i.e.Ψ D 2n+1 (m)
is invariantunderthesignhanges. Now
m = x i 1 1 . . . x i 2n+1 2n+1 y 1 j 1 . . . y 2n+1 j 2n+1
withP 2n+1
k=1 (i k + j k ) = 2p
,therefore atleastoneofthei k + j k
iseven.Let'sdenoteby
l
theorrespondingindex.So,forevery
k 6= l
,wehaves k (m) = (−1) i k +j k m = s k,l (m)
ands l (m) = m
.ButΨ 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
sik = l
= Ψ D 2n+1 (m)
.•
SowehaveS 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}
.•
ForP ∈ S B sl n 2
,equation(17)maybewritten2n+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
.Let
P
beinvariantbysignhanges.IfP
issolution ofequation (11)forD n
,thenP
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
)bethegroupofsignhangesofB n
(resp.D n
).WemaywriteSB n = SD n ⊔ SD n · s 1
.Let
P
beinvariantbysignhanges.SowehaveP = R B n (P ) = R D n (P)
,andequation(11)forB n
(resp.D n
)maybewritten
E B n (P ) = R B n (Q)
(resp.E n D (P ) = R D n (Q)
),withQ = (z · y − t · x) P(x + z y + t)
.If
P
issolutionofequation(11)forD 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)forB 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)inthespaeS 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
,thedimensionofHH k (A n (C) W )
isthenumberofonjugaylassesofW
admittingtheeigenvalue1
withthemultipliityk
.Byspeializingtotheasesof
B n
andD n
,weobtain:Corollary 19 (Alev-Farinati-Lambre-Solotar)
•
FortypeB n
,thedimensionofHH 0 (A n (C) W )
isthenumberof partitionsπ(n)
of theintegern
.•
FortypeD n
, thedimensionofHH 0 (A n (C) W )
isthenumberofpartitionse π(n)
oftheintegern
havinganevennumber ofparts.
TheonjetureofJ.Alevmaybesetforthasfollows :
Conjeture 20 (Alev)
•
ForthetypeB n
,thedimensionofHP 0 (S W )
equalsthenumberofpartitionsπ(n)
oftheintegern
.•
Forthe typeD n
, the dimensionofHP 0 (S W )
equals thenumber of partitionse π(n)
of theintegern
having anevennumberofparts.
Now,letusshowhowtoonstrut
π(n)
solutionsofequation(11)fortheaseofB n
.Denition 21
For
i 6= j
, wenoteX 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)graphg Γ M
suh that⊲
theset ofvertiesofg Γ M
istheset ofindies{k ∈ [[1, n]] / ∃ i ∈ [[1, n]] / a i,k 6= 0
ora k,i 6= 0}
,⊲
twovertiesi, j
ofΓ g M
areonneted bytheedgei
a i,j
j
ifa i,j 6= 0
.•
Ifσ ∈ S n
,thenthegraphΓ ] σ·M
isobtainedbypermutingthevertiesofg Γ M
.So,byreplaingeahvertexbythesymbol
•
,weobtainagraphΓ M
suhthatthemapM 7→ Γ M
isonstantonev-eryorbitundertheationof
B n
(resp.D n
).SowemayassoiatethisgraphtotheelementR n Q n−1 i=1
Q n
j=i+1 X i,j 2a i,j
.