Finite Dimensional Lie Algebras and their Representations [Lecture notes]

49 Pages·2012·0.437 MB·English

by  Ian Grojnowski, with C. Almost

Finite Dimensional Lie Algebras and
their Representations
Michaelmas 2006
∗
Prof. I. Grojnowski
ChrisAlmost†
Theexamquestionswillconsistofasubsetofthein-classexercises.
Contents
Contents 1
1 MotivationandSomeDefinitions 2
1.1 LieGroupsandLieAlgebras . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Representationsofsl 7
2
2.1 Classification. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Semi-SimpleLieAlgebras: StructureandClassification 12
3.1 SolvableandNilpotentLieAlgebras . . . . . . . . . . . . . . . . . . . 13
3.2 StructureTheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4 RootSystems 22
4.1 AbstractRootSystems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.2 DynkinDiagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.3 ExistenceandUniquenessofanAssociatedLieAlgebra. . . . . . . . 28
5 RepresentationTheory 30
5.1 TheoremoftheHighestWeight . . . . . . . . . . . . . . . . . . . . . . 30
5.2 ProofoftheTheoremoftheHighestWeight . . . . . . . . . . . . . . 33
5.3 WeylCharacterFormula . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6 Crystals 41
6.1 LittelmannPaths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Index 49
∗[email protected]
†[email protected]
1
2 LieAlgebras
1 Motivation and Some Definitions
1.1 Lie Groups and Lie Algebras
This class is about the structure and representations of the so-called simple Lie
groups
SL ={A∈Mat |detA=1},SO ={A∈SL |AAT =I},Sp
n n n n 2n
and exactly five others. The only technology we will use is linear algebra. These
groupsareslightlycomplicated,andtheyhaveatopologyonthem. Tostudythem
wepasstoa“linearization,”theLiealgebra,andstudytheseinstead.
1.1.1Definition. A linear algebraic group is a subgroup of GL , for some n, de-
n
finedbypolynomialequationsinthematrixcoefficients.
1.1.2Example. Allofthegroupsmentionedsofararelinearalgebraicgroups,as
isthecollectionofuppertriangularmatrices.
This definition requires an embedding; it is desirable to come up with one
whichdoesnot. Itturnsoutthelinearalgebraicgroupsareexactlytheaffinealge-
braic groups, i.e. affine algebraic varieties such that multiplication and inversion
areregularfunctions.
1.1.3Definition. TheLiealgebraassociatedwithG isg=T G,thetangentspace
1
toG attheidentity.
1.1.4Example. Let g=(cid:2)10(cid:3)+(cid:34)(cid:148)a b(cid:151),where(cid:34)2=0. Then
01 c d
detg=(1+(cid:34)a)(1+(cid:34)d)−bc(cid:34)2=1+(cid:34)(a+d)
Thereforedetg=1ifandonlyiftrg=0.
We are working in the ring E =(cid:67)[(cid:34)]/(cid:34)2 =0 (the so-called dual numbers). If
G⊆GL isalinearalgebraicgroup,wecanconsider
n
G(E)={A∈Mat (E)|AsatisfiestheequationsdefiningG}.
n
There is a natural map π : G(E) → G((cid:67)) induced by the ring homomorphism
π:(cid:34)(cid:55)→0.
1.1.5Definition. G(E)isthetangentbundletoG,denoted TG,so
g=T G=π−1(I)={A∈Mat (cid:67)|I+(cid:34)A∈G(E)}.
1 n
1.1.6Exercise. Show that this definition is equivalent to the usual definition of
TG (say,fromdifferentialgeometry).
1.1.7Examples.
1. LetG=GL ={A∈Mat |A−1∈Mat }. Then
n n n
G(E)=GL ⊕Mat (cid:34)={A+B(cid:34)|A−1∈Mat },
n n n
since(A+B(cid:34))(A−1−A−1BA−1(cid:34))=I. Therefore
gl :=T GL =Mat ∼=(cid:82)n2.
n 1 n n
LieGroupsandLieAlgebras 3
2. As an exercise, show that det(I +X(cid:34)) = 1+tr(X)(cid:34). As a corollary, sl :=
n
T SL ={X ∈Mat |tr(X)=0}.
1 n n
3. Let G = O = {A| ATA= I}. Then (I +X(cid:34))T(I +X(cid:34))= I +(XT +X)(cid:34), so
n
o :=T O ={X ∈Mat ((cid:67))|XT +X =0}.
n 1 n n
Remark. IfX ∈o thentrX =0(when2(cid:54)=0),sotheLiealgebraofO iscontained
n n
intheLiealgebraofSL . Alsonotethato =so ,sotheLiealgebradoesnotkeep
n n n
trackofwhetherthegroupisconnected.
What structure on g is there coming from the group structure on G? Notice
that
(I+A(cid:34))(I+B(cid:34))=I+(A+B)(cid:34),
so traditional multiplication only sees the vector space structure. Instead we use
thecommutatormap
G×G→G:(P,Q)(cid:55)→PQP−1Q−1,
and consider it infinitesimally. If P = I+A(cid:34) andQ = I+Bδ, then P−1 = I−A(cid:34)
andQ−1=I−Bδ,so
PQP−1Q−1=I+(AB−BA)(cid:34)δ.
Let[A,B]denotethemap(A,B)(cid:55)→AB−BA.
1.1.8Exercises.
1. Showthat(PQP−1Q−1)−1=QPQ−1P−1 implies[X,Y]=−[Y,X]
2. ShowthatthefactthatmultiplicationisassociativeinG implies
[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0
1.1.9Definition. LetK beafieldofcharacteristicnot2or3. ALiealgebragover
K is a vector space over K equipped with a bilinear map [·,·]:Λ2g→g, the Lie
bracket,suchthatforallX,Y,Z ∈G,
1. [X,Y]=−[Y,X](skew-symmetry);and
2. [[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0(Jacobiidentity).
1.1.10Example. In most of the following the Lie bracket is inherited from gl .
n
ThefollowingareLiealgebras.
1. sl ={A∈gl |trA=0};
n n
2. so ={A∈gl |AT +A=0};
n n
3. sp ={A∈gl |JATJ−1+A=0},where
2n 2n
J =anti-diag(−1,...,−1,1,...,1).
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
n n
4. Uppertriangularmatrices;
4 LieAlgebras
5. Strictlyuppertriangularmatrices;
6. Let V be any vector space and define the Lie bracket to be the zero map.
ThisexampleisknownastheAbelianLiealgebra.
1.1.11Exercises.
1. Check directly that gl = Mat is a Lie algebra with Lie bracket [A,B] =
n n
AB−BA,andinfactthisworksforanyK-algebra;
2. Check that all of the above are truly subalgebras, i.e. vector subspaces and
closedundertheLiebracket.
1.1.12Exercise. ClassifyallLiealgebrasgoverK withdim g≤3.
K
1.2 Representations
1.2.1Definition. A Lie algebra homomorphism ϕ : g → g is a linear map such
1 2
that
ϕ([x,y] )=[ϕ(x),ϕ(y)] .
1 2
ALiealgebrarepresentation(orsimplyrepresentation)ofaLiealgebragonavector
spaceV isaLiealgebrahomomorphismϕ:g→gl .
V
1.2.2Example. If g ⊆ gl then the inclusion is a representation of g on V. For
V
example,Kn isarepresentationofso ⊆gl .
n n
1.2.3Definition. Let x ∈ganddefinetheadjointof x tobe
adx :g→g: y (cid:55)→[x,y],
alinearmapong.
1.2.4Lemma. The adjoint map ad : g → End(g) is linear and is in fact ad is a
representationofgonitselfcalledtheadjointrepresentation.
PROOF: Wemustcheckthatad[x,y]=adxady−adyadx. Foranyz,
0=[[x,y],z]+[[y,z],x]+[[z,x],y] Jacobiidentity
0=[[x,y],z]−[x,[y,z]]+[y,[x,z]] skew-symmetry
[[x,y],z]=[x,[y,z]]−[y,[x,z]]
soad[x,y](z)=(adxady−adyadx)(z)foreveryz,implyingtheresult. (cid:131)
1.2.5Definition. Thecenterofgis
Z ={x ∈g|[x,y]=0forevery y ∈g}=ker(ad:g→End(g)).
g
Warning: the notation Z is not used in class (in fact, no notation for center is
g
usedinclass).
Bythedefinitionofcenter,theadjointhomomorphismembedsgintogl ifand
g
onlyifghastrivialcenter.
Representations 5
1.2.6Example. Theadjointhomomorphismmapsg=(cid:2)0K(cid:3)tozeroingl .
0 0 g
1.2.7Theorem(Ado). Any finite dimensional Lie algebra g has a finite dimen-
sionalfaithfulrepresentation,i.e.g(cid:44)→gl forsomen.
n
PROOF: Omitted. (cid:131)
1.2.8Example. Recallthatsl ={(cid:2)a b (cid:3)}. Itsstandardbasisis
2 c −a
(cid:26) (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)(cid:27)
0 1 0 0 1 0
e= ,f = ,h= .
0 0 1 0 0 −1
Showthat[h,e]=2e,[h,f]=−2f,and[e,f]=h. Whence,arepresentationof
sl on(cid:67)n isatripleofmatrices E,H,F suchthat[H,E]=2E,[H,F]=−2F,and
2
[E,F]=H. Howdowefindsuchathing?
1.2.9Definition. If G isanalgebraicgroupthenanalgebraicrepresentationof G
isagrouphomomorphismρ:G→GL definedbypolynomialequations.
V
We pull the same trick and substitute E = (cid:67)[(cid:34)]/(cid:34)2 for (cid:67) to get a homomor-
phismofgroupsG(E)→GL (E). Asρ(I)=I,
V
ρ(I+A(cid:34))=I+(cid:34)(somefunctionofA).
Callthisfunctiondρ,sothatρ(I+A(cid:34))=I+(cid:34)dρ(A),givingdρ:g→gl .
V
1.2.10Exercises.
1. Showthatifρ:G→GL isagrouphomomorphismthendρ:g→gl isa
V V
Liealgebrahomomorphism.
2. Showthat dρ trulyisthedifferentialofρ attheidentitywhenρ isconsid-
eredasasmoothmapbetweenmanifolds.
We have a functor ρ (cid:55)→ dρ from the algebraic representations of G to the
Lie algebra representations of g = T G. We will discuss this further, but first an
1
example.
1.2.11Example. Let G = SL , and let L(n)1 be the collection of homogeneous
2
polynomialsofdegreeninvariables x and y,sodim L(n)=n+1. Indeed,ithas
K
abasis{xn,xn−1y,...,xyn−1,yn}. NowGL actson L(n)via
2
ρn:GL2→GL(L(n))=GLn+1
by(ρ ((cid:148)a b(cid:151))f)(x,y)= f(ax+cy,bx+dy). Thenρ isthetrivialrepresenta-
n c d 0
tion, ρ is the standard 2-dimensional representation (i.e. ρ is the identity map
onGL1),andρ (cid:148)a b(cid:151)hasmatrix 1
2 2 c d
 
a2 ab b2
2ac ad+bc 2bd.
 
c2 cd d2
1L(n)=Γ((cid:80)1,(cid:79)(n))andH1((cid:80)1,(cid:79)(n))=0foralln≥0.
6 LieAlgebras
Now of course SL < GL so the restriction is a representation of SL . To
2 2 2
computedρ (e)(xiyj)(recalle=(cid:2)01(cid:3))wecompute:
n 00
ρ (1+(cid:34)e)xiyj =ρ (cid:2)1(cid:34)(cid:3)xiyj =xi((cid:34)x+y)j =xiyj+(cid:34)jxi+1yj−1
n n 01
Thereforedρ (e)(xiyj)= jxi+1yj−1. Similarly,
n
ρ (1+(cid:34)f)xiyj =ρ (cid:2)10(cid:3)xiyj =(x+(cid:34)y)iyj =xiyj+(cid:34)ixi−1yj+1
n n (cid:34) 1
and
ρ (1+(cid:34)f)xiyj =ρ (cid:148)1+(cid:34) 0 (cid:151)xiyj =(1+(cid:34))ixi(1−(cid:34))jyj =xiyj+(cid:34)(i−j)xiyj,
n n 0 1−(cid:34)
so
e·xiyj = jxi+1yj−1 f ·xiyj =ixi−1yj+1 h·xiyj =(i−j)xiyj.
Uponcloserinspection,theseareoperatorsthatweknowandlove,
∂ ∂ ∂ ∂
dρ (e)=x , dρ (f)= y , dρ (h)=x −y . (1)
n ∂ y n ∂x n ∂x ∂ y
1.2.12Exercises.
1. Checkdirectlythatdefiningtheactionof e, f,andhon L(2)by(1)givesa
representationofsl .
2
2. CheckthatL(2)istheadjointrepresentationofsl (noticethattheyareboth
2
3-dimensional).
3. Ifthecharacteristicof K iszero,checkthat L(n)isanirreduciblerepresen-
tationofsl ,andhenceofthegroupSL .
2 2
1.2.13Example. Let’s compare representations of G and representations of g
when G = (cid:67)∗ (and hence g = (cid:67) with [x,y] = 0 for all x,y ∈ (cid:67)). Recall that
theirreduciblealgebraicrepresentationsofG areonedimensional. z∈(cid:67)∗ actsas
multiplication by zn (n ∈ (cid:90)). Moreover, any representation of G splits up into a
finitedirectsumofonedimensionalrepresentations(“finiteFourierseries”).
In contrast, a representation ρ : g → End(V) of g = (cid:67) on V is any matrix
A = ρ(1) ∈ End(V) (ρ : λ (cid:55)→ λA). A subrepresentation W ⊆ V is just an A-
stable subspace (so AW ⊆ W). Any linear transformation has an eigenvector,
so there is always a 1-dimensional subrepresentation, which implies that all ir-
reducible representations are the 1-dimensional ones. V breaks up into a direct
sum of irreducibles (i.e. is completely reducible) if and only if A is diagonaliz-
able. e.g. if A is the Jordan block of size n, then the invariant subspaces are
〈e 〉,〈e ,e 〉,...,〈e ,...,e 〉,anddoesnotbreakupintoadirectsumofsubspaces.
1 1 2 1 n
Representationsuptoisomorphismofgl aregivenbythetheoryofJordannormal
1
forms.
Representations of G and g look “really different” in this case. Even the
irreducibles are different, since for G they are in 1-1 correspondence with (cid:90)
(z (cid:55)→ zn), while for g they are in 1-1 correspondence with (cid:67) (λ (cid:55)→ “1 (cid:55)→ λ”).
(Here(cid:90)(cid:44)→(cid:67):z(cid:55)→2πiz.)
Representationsofsl 7
2
In contrast, the map from representations of G to representation of g is an
equivalenceofcategoriesifG isasimplyconnectedsimplealgebraicgroup. Inpar-
ticular, every representation of the Lie algebra sl is a direct sum of irreducibles,
2
andtheirreduciblesareexactly L(n).
2 Representations of sl
2
2.1 Classification
From now on, all representations and Lie algebras are over (cid:67) unless otherwise
specified.
2.1.1Theorem.
1. Thereisauniqueirreduciblerepresentationofsl ofdimension(n+1)for
2
everyn≥0.
2. Everyfinitedimensionalrepresentationofsl isisomorphictoadirectsum
2
ofirreduciblerepresentations.
PROOF(FIRST PART): LetV bearepresentationofsl2. Define
Vλ:={v∈V |hv=λv},
the λ-weight space. It is the space of eigenvectors of h with eigenvalue λ. For
example, L(n)λ=(cid:67)xiyj ifi−j=λ,andinparticular L(n)n=(cid:67)xn. Supposethat
v∈Vλ,andconsiderev.
h(ev)=(he−eh+eh)v=[h,e]v+ehv=2ev+λev=(2+λ)ev
Whence v∈Vλ ifandonlyifev∈Vλ+2. Similarly, v∈Vλ ifandonlyif fv∈Vλ−2.
A highest weight vector of weight λ is an element of ker(e)∩Vλ, i.e. a v ∈ V
suchthatev=0andhv=λv.
Claim. If v is a highest weight vector then W = span{v,fv,f2v,...} is an sl -
2
invariantsubspaceofV.
We need to show that W is stable by e,f,h. It is clear that fW ⊆ W. As
hfkv=(λ−2k)fkv,itisclearthathW ⊆W. Bydefinition,ev=0,so
efv=(ef −fe)v+fev=hv=λv∈W.
Similarly,
efk+1v=(ef −fe+fe)fkv
=hfkv+fefkv
=(λ−2k)fkv+k(λ−k+1)fkv byinduction
=(k+1)(λ−k)fkv,
so efkv = k(λ−k+1)fk−1v ∈ W, by induction, so eW ⊆ W and the claim is
proved.
8 LieAlgebras
Claim. If v is a highest weight vector of weight λ then λ ∈ {0,1,2,...} if V is
finitedimensional.
Indeed, the vectors fkv all lie in different eigenspaces of h, and hence are
linearlyindependentiftheyarenon-zero. If V isfinitedimensionalthen fkv=0
forsomek. Letkbeminimalwiththisproperty. Then
0=efkv=k(λ−k+1)fk−1v,
andk≥1,soλ=k−1∈{0,1,2,...}since fk−1v(cid:54)=0.
Claim. ThereexistsahighestweightvectorifV isfinitedimensional.
Letv∈V beanyeigenvectorforhwitheigenvalueλ(suchthingsexistbecause
(cid:67) is algebraically closed). As before, v,ev,e2v,... are all eigenvectors for hwith
eigenvalues λ,λ+2,λ+4,.... Hence they are linearly independent, so there is
some k+1minimalsuchthat ek+1v =0. Clearly ekv ∈Vλ+2k isahighestweight
vector,provingtheclaim.
If V is irreducible and dim(cid:67)V = n+1 then there is a basis {v0,...,vn} such
that
hvi =(n−2i)vi, fvi =(cid:168)0vi+1 iiffii<=nn, evi =i(n−i+1)vi−1.
Indeed,let v beahighestweightvectorwithweightλ∈{0,1,2,...}. Thestring
0
{v,fv,...,fλv}isasubrepresentationofV. AsV isirreducibleitmustbeallofV,
soλ=dim(cid:67)V −1. Take vi = fiv. Thisprovesthefirstpartofthetheorem. (cid:131)
2.1.2Exercise. (cid:67)[x,y]isarepresentationofsl bytheformulaein1.2.11. Show
2
thatforλ,µ∈(cid:67), xλyµ(cid:67)[x,y]isalsoarepresentation,infinitedimensional,and
analyzeitssubmodulestructure.
PROOF(SECOND PART): Wewillshowthatstringsofdifferentlengths“donotinter-
act,”andthenthatstringsofthesamelength“donotinteract.”
LetV beafinitedimensionalrepresentationofsl . Let
2
Ω:=ef +fe+1h2∈End(V),
2
theCasimirofsl . AlsonoticethatΩ=(1h2+h)+2fe.
2 2
Claim. (1)Ωiscentral,i.e.Ωe=eΩ,Ωf = fΩ,andΩh=hΩ.
Theproofofthisisanexercise. Forexample,
Ωe=e(ef +fe+1h2)
2
=e(ef −fe)+2efe+1heh+1(eh−he)h
2 2
=eh+2efe+1heh−eh
2
=2efe+1heh
2
=(ef +fe+1h2)e
2
Classification 9
Claim. (2)IfV isanirreduciblerepresentationwithhighestweightnthenΩacts
onV bymultiplicationby 1n2+n.
2
(ItfollowsfromSchur’sLemmathatifV isanirreduciblerepresentationofsl
2
then Ω acts on V as multiplication by a scalar.) Let v be a highest weight vector,
sohv=nv andev=0,so
Ωv=((1h2+h)+2fe)v=(1n2+n)v.
2 2
Hence Ω(fkv)= fk(Ωv)=(1n2+n)fkv, so Ω acts by multiplication by 1n2+n
2 2
since{v,fv,...,fnv}isabasisforV.
Claim. (3) If L(n) and L(m) are both irreducible representations and Ω acts on
∼
thembythesamescalarωthenn=m(so L(n)=L(m))andω= 1n2+n.
2
Indeed, L(n) and L(m) are irreducible representations with highest weight n
andm,respectively,sobythelastclaim 1n2+n=ω= 1m2+m. But f(x)= 1x2+
2 2 2
x is a strictly increasing function for x > −1, so m = n and the representations
areisomorphic.
Let Vω = {v ∈ V | (Ω−ωI)dim(cid:67)Vv = 0} be the generalized eigenspace of Ω
associatedwithω. TheJordandecompositionforΩimpliesthat V ∼=(cid:76) Vω (as
ω
vectorspaces).
ω
Claim. (4) Each V is a subrepresentation of V, i.e. the above is a direct sum
decompositionasrepresentationsofsl .
2
Indeed,if x ∈sl and v∈Vω thensinceΩiscentral,
2
(Ω−ω)dim(cid:67)Vxv=x(Ω−ω)dim(cid:67)Vv=0.
Aside: Forag-moduleW,acompositionseriesisasequenceofsubmodules
0=W <W <···<W =W
0 1 r
suchthateachquotientmoduleWi/Wi−1isanirreducibleg-module. Forexample,
withg=(cid:67)andW =(cid:67)2,actingas1(cid:55)→(cid:2)01(cid:3),then〈e 〉⊆〈e ,e 〉isacomposition
00 1 1 2
series (and this is the only one). If the action is instead 1(cid:55)→(cid:2)00(cid:3) then any line
00
in(cid:67)2 givesrisetoacompositionseries.
RecallthatifW isfinitedimensionalthenithasacompositionseries,andthe
subquotientsareuniqueuptopermutations.
Claim. (5)If Vω (cid:54)=0thenω= 1n2+nforsome n,andallthesubquotientsofa
compositionseriesfor Vω are L(2n), i.e. Vω isabunchofcopiesof L(n), possibly
stucktogetherinacomplicatedway.
Indeed,supposeW isanirreduciblesubrepresentationof Vω. ThenΩactson
W asmultiplicationbyω,sinceωistheonlypossiblegeneralizedeigenvalueand
Ω acts on irreducible representations as scalar mulitplication by Schur’s Lemma.
∼
Thusω= 1n2+nforsomenwithW =L(n),byclaim(3). Bythesamereasoning,
Ωwillacto2nanyirreduciblesubrepresentationof Vω/W asmultiplicationbyω,
soanysuchirreduciblesubrepresentationmustbe L(n), forthesame n. Whence
everysubquotientinacompositionseriesisisomorphicto L(n).
10 LieAlgebras
Tofinishtheproofwemustshowthateach Vω isadirectsumof L(n)’s. The
keypointwillbetoshowthathis“diagonalizable.” Atthisstagewemayassume
thatV =Vω.
Noticethathactson V witheigenvalues{n,n−2,...,2−n,−n}. (Itisafact
fromlinearalgebrathatifhactsonavectorspaceW andW(cid:48)≤W isanh-invariant
(cid:48)
subspacethentheeigenvaluesofhonW aretheeigenvaluesofhonW together
with the eigenvalues of hon W/W(cid:48). The remark follows from this since V has a
composition series with subquotients all isomorphic to L(n), and we know from
thefirstpartthattheeigenvaluesofhon L(n)are{n,n−2,...,2−n,−n}.) Hence
V =0if m∈/ {n,n−2,...,2−n,−n}. Itfollowsthathhasonlyonegeneralized
m
Whydoesthisfollow? eigenvalueonker(e),namelyn.
Claim. hfk= fk(h−2k)andefn+1= fn+1e+(n+1)fn(h−n).
Thefirstequationistrivial, andthesecondisprovedbyinductionasfollows.
Clearlyef = fe+ef −fe= fe+h,and
efn+1=(fe+ef −fe)fn
= fn+1e+nfn(h−n+1)+fn(h−2n) byinduction
= fn+1e+(n+1)fn(h−n).
Claim. hactsasmultiplicationbynonker(e),andfurther,V =ker(e).
n
Indeed, if v ∈ Vn (so hv = nv) then ev ∈ Vn+2 = 0, so Vn ⊆ ker(e). For the
converse,considerthesestatements:
1. If x ∈ ker(e), then (h−n)dimVx = 0 since n is the only generalized eigen-
valueofhonker(e).
2. If x ∈ker(e),then(h−n+2k)dimVfkx = fk(h−n)dimVx =0bytheprevious
claim.
3. If y ∈ker(e) and y (cid:54)=0 then fny (cid:54)=0. (Indeed, let (W) be a composition
i
series for V, as abo∼ve. Then there is i such that y ∈ Wi \Wi−1. Let ¯y =
y +Wi ∈ Wi/Wi−1 = L(n). Then ¯y is a highest weight vector of L(n), so
fn¯y (cid:54)=0,hence fny (cid:54)=0.)
4. fn+1x =0,as fn+1xisinthegeneralizedeigenspacewitheigenvalue−n−2,
whichiszero.
Hence
0=efn+1x =(n+1)fn(h−n)x+fn+1ex =(n+1)fn(h−n)x
implying fn((h−n)x)=0,sobystatement3,(h−n)x =0,orhx =nx.
Finally,chooseabasis{w1,...,w(cid:96)}ofVn=ker(e). Then
span{w ,fw ,...,fnw }⊕span{w ,fw ,...,fnw }⊕...
1 1 1 2 2 2
⊕span{w(cid:96),fw(cid:96),...,fnw(cid:96)}
is a direct sum decomposition of V into subrepresentations, each isomorphic to
L(n). (cid:131)