An introduction to Lie algebras and the theorem of Ado

23 Pages·2013·0.28 MB·English

An introduction to Lie algebras and the theorem of Ado
Hofer Joachim
28.01.2012
AnintroductiontoLiealgebrasandthetheoremofAdo
Contents
Introduction 2
1 Liealgebras 3
1.1 Subalgebras,ideals,quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Nilpotent,solvable,simpleandsemisimpleLiealgebras . . . . . . . . . . . . . . . . . . . . . . . 5
2 TherepresentationtheoryofLiealgebras 7
2.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Modules,submodules,quotientmodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 Structuretheorems: LieandEngel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3 ThetheoremofAdofornilpotentLiealgebras. Howcanafaithfulrepresentationbeconstructed? 11
3.1 Adhocexamples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Theuniversalenvelopingalgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2.1 Tensorproductsandthetensoralgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2.2 TheuniversalenvelopingalgebraofaLiealgebra . . . . . . . . . . . . . . . . . . . . . . 12
3.2.3 ThePoincare-Birkhoff-Witttheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.3 Constructingafaithfulrepresentationofh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1
3.4 Ado’stheoremfornilpotentLiealgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.5 ConstructingafaithfulrepresentationforthestandardfiliformLiealgebraofdimension4 . . . . . 14
3.6 ConstructingafaithfulrepresentationforanabelianLiealgebra . . . . . . . . . . . . . . . . . . . 15
4 ThetheoremofAdo 16
4.1 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4.2 DirectandsemidirectsumsofLiealgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4.3 ProofofAdo’stheorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.4 ProofofNeretin’slemma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4.5 Constructingafaithfulrepresentationofthe2-dimensionaluppertriangularmatrices . . . . . . . 19
4.6 ConstructingafaithfulrepresentationofanabstractLiealgebra . . . . . . . . . . . . . . . . . . . 19
1 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
Introduction
LiegroupsandLiealgebrasareofgreatimportanceinmodernphysics,particularlyinthecontextof(continu-
ous) symmetry transformations. The Lie algebra of aLie group is defined as the tangent space to the neutral
element of the group and its elements can be seen as "infinitesimal transformations". The Lie algebra of a
Lie group is uniquely determined (the converse is not true unless the group is simply connected) and many
questions about the group can be reduced to questions about the Lie algebra, which are usually easier to
handle. Itisparticularlypleasantifthealgebracanberepresentedbymatricesandanimportantresultinthis
area is given by Ado’s theorem, which states that any finite-dimensional Lie algebra can be represented by
(finite) matrices. In this thesis we will prove Ado’s theorem for nilpotent Lie algebras and provide a method to
construct such matrix representations. It is also worthwhile to mention that, although Lie algebras historically
aroseasameanstostudyLiegroups,theyaremeanwhileoftenstudiedintheirownright.
The first chapter contains the basic definitions and some helpful examples. In the second chapter a short
introductiontorepresentationtheoryisgivenaswellastheproofstoEngel’stheoremandLie’stheorem. The
third chapter is reserved for the proof of Ado’s theorem for nilpotent Lie algebras and the theory needed for it
(also,theexplicitconstructionoffaithfulrepresentationsisshownintwoexamples). Finally,thefourthchapter
containstheprooftoAdo’stheoremforarbitraryLiealgebrasaswellastheneededtheory.
TheproofstoEngel’sandLie’stheoremsare,forthemostpart,basedontheproofsgivenin[1]. Theproofof
Ado’stheoremfornilpotentLiealgebras(section3)isthesameasgivenin[3],theproofofAdo’stheoremfor
arbitraryLiealgebrasisbasedontheonegivenin[5].
In the following, with the exception of the construction of the universal enveloping algebra in chapter 3.2.,
allvectorspacesareassumedtobefinite-dimensional. Furthermore,unlessmentionedotherwise,theunder-
lyingfieldsareofcharacteristiczeroandalgebraicallyclosed. NotehoweverthatAdo’stheoremisvalidforLie
algebrasoverfieldsofarbitrarycharacteristic(and,inthiscontext,issometimescalledIwasawa’stheorem).
2 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
1 Lie algebras
Definition. A Lie algebra over a field F is a vector space L over F, together with an operation L×L → L,
(x,y)→[x,y],whichfulfillsthefollowingaxioms:
(i) Theoperation[.,.]isbilinear.
(ii) ∀x∈L: [x,x]=0.
(iii) ∀x,y,z ∈L: [x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0.
The operation [.,.] is usually called the bracket operation. Property (iii) is called the Jacobi identity. Note
that(ii)implies
(ii(cid:48)) ∀x,y ∈L: [x,y]=−[y,x].
IfcharF(cid:54)=2,(ii)and(ii(cid:48))areequivalent.
An important example for a Lie agebra is the set of all linear transformation V → V (where V is any vec-
tor space of a field F), denoted by EndV or, in the context of Lie algebras, gl(V). gl(V) is itself a vector
spacewithdimension(dimV)2 andaringw.r.t. thecompositionofmaps. Thebracketoperationisdefinedby
[x,y]=xy−yx∀x,y ∈gl(V),wherexyisthecompositionofxandy. AfterabasisofV hasbeenchosen,the
elements of gl(V) can be represented as n×n matrices and we write gl (F). The standard basis consists of
n
matricese ,havingaoneinthe(i,j)positionandzeroseverywhereelse. TheLiebracketisthengivenby
ij
[e ,e ]=δ e −δ e .
ij kl jk il il jk
Every vector space V with the bracket operation defined as [x,y] = 0 ∀x,y ∈ V is a Lie algebra. Such Lie
algebrasarecalledabelian.
If L is one-dimensional, it has exactly one basis vector x with the commutation relation [x,x] = 0, so any
one-dimensionalLiealgebraisabelien.
IfListwo-dimensionalwithbasisvectorsx,y,itiseitherabelienor
[x,y]=αx+βy.
Nowdefineanewbasisbyx(cid:48) =αx+βy andtakey(cid:48) tobeanorthogonalvector. Itfollowsthat
[x(cid:48),y(cid:48)]=γx(cid:48)
andbyscalingy(cid:48) →γ−1y(cid:48) weget
[x(cid:48),y(cid:48)]=x(cid:48).
Therefore,uptoisomorphism,thereareexactlytwotwo-dimensionalLiealgebras,oneofwhichisabelian.
The Lie algebra sl (R) is the vector space of all real 2 × 2-matrices with trace zero with the commutator
2
asLiebracket. Itisspannedbythematrices
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
0 1 0 0 1 0
e= , f = , g = .
0 0 1 0 0 −1
Thecommutationrelationsare
[e,f]=g, [e,g]=−2e, [f,g]=2f.
Moregenerally,sl (R)/sl (C)isthespaceofallreal/complexn×n-matriceswithtracezero.
n n
TheHeisenbergLiealgebrah isa(2n+1)-dimensionalrealvectorspacewithabasis{x , ..., x , y , ..., y , z}
n 1 n 1 n
andtheLiebracketdefinedby
[x ,y ]=z
i i
andallotherbracketsequaltozero. Forexample,h canbeidentifiedwith(i.e. "afaithfulrepresentationofh
1 1
isgivenby",seethedefinitionsbelow)thespaceofrealmatricesspannedby
3 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
     
0 1 0 0 0 0 0 0 1
x=0 0 0, y =0 0 1, z =0 0 0
0 0 0 0 0 0 0 0 0
andtheLiebracketisnowgivenbythecommutator.
The following example shows the connection between a Lie algebra and the corresponding Lie group (by a
naiveapproach). LetAff(R)bethegroupofallinvertibleaffinetransformationsoftheline,i.e.
Aff(R)={L :R→R|L x=ax+b, a,b∈R, a(cid:54)=0}.
a,b a,b
TheelementsofAff(R)canbewrittenasmatricesoftheform
(cid:18) (cid:19)
a b
L = ,a(cid:54)=0
a,b 0 1
(cid:18) (cid:19)
x
and the transformation is given by their action on vectors of the form . Now let l be a 2×2-matrix, such
1
that1+(cid:15)l∈Aff(R)(where1denotestheidentitymatrixand(cid:15)∈R). Obviouslylhastobeoftheform
(cid:18) (cid:19)
a b
l= .
0 0
TheseelementsformtheLiealgebraaff(R)(withthecommutatorastheLiebracket). Notethatingeneralone
has to use the exponential map to get from the Lie algebra to the Lie group (and to define the Lie algebra of
theLiegroup). Anotherdownsideofthisapproachisthatitdoesn’texplainhowthegroupmultiplicationleads
totheLiebracket.
TheaffinegroupAff(Rn)isthegroupofallinvertibleaffinetransformationsRn →Rn. Asabove,itisformed
byelementsoftheform
(cid:18) (cid:19)
A b
L = ,A∈GL (R),b∈Rn
A,b 0 1 n
andthecorrespondingLiealgebraisgivenby
(cid:18) (cid:19)
A b
aff(Rn)={ |A∈gl (R), b∈Rn}.
0 0 n
Definition. A Lie algebra homomorphism between two Lie algebras L and L is a linear map φ : L → L ,
1 2 1 2
suchthat
∀x,y ∈L : φ([x,y])=[φ(x),φ(y)].
1
1.1 Subalgebras, ideals, quotients
Definition. AsubspaceK ofaLiealgebraLiscalledaLiesubalgebra,if
x,y ∈K =⇒ [x,y]∈K.
Obviously, every one-dimensional subspace of a Lie algebra is a subalgebra. The Lie algebra sl (C) is a
2
subalgebraofgl (C).
2
ThenormalizerofasubspaceK ofLisdefinedas
N (K)={x∈L|[x,K]⊂K}.
L
N (K)isasubalgebraofL,becauseforx∈K andx , x ∈N (K),theJacobiidentityimplies
L 1 2 L
(cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3)
[x ,x ],x = x ,[x ,x] − x ,[x ,x]
1 2 1 2 2 1
andtherefore[x ,x ] ∈ N (K). IfK isasubalgebra, N (K)isthelargestsubalgebraof LwhichincludesK
1 2 L L
asanideal.
Definition. AsubspaceI ofaLiealgebraLiscalledanideal,if
∀x∈L∀i∈I : [x,i]∈I.
4 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
Obviously every ideal is also a subalgebra. An example for an ideal is the center Z(L) = {x ∈ L|[x,l] =
0∀l∈L}. IfZ(L)=L,Liscalledabelian. Anotherexampleis[L,L]={[l ,l ]|l ,l ∈L}.
1 2 1 2
Consider the Heisenberg Lie algebra h with the basis {x , ..., x , y , ..., y , z} and the Lie bracket de-
n 1 n 1 n
finedby[x ,y ]=z. Thentheone-dimensionalsubspacespannedbyz isanidealofh ,namely[h ,h ]=(cid:104)z(cid:105).
i i n n n
Proposition 1. The kernel Ker(φ) of a Lie algebra homomorphism φ : L → L is an ideal of L , its image
1 2 1
Im(φ)isasubalgebraofL .
2
Proof. That the kernel and the image are subspaces is a simple consequence of the linearity of homomor-
phisms. Since
[φ(x),φ(y)]=φ([x,y]),
Im(φ)isasubalgebra. Nowassumex∈Ker(φ). Then,foreveryy ∈L ,
1
φ([x,y])=[φ(x),φ(y)]=[0,φ(y)]=0
andtherefore[x,y]∈Ker(φ).
Definition. Given a Lie algebra L and an ideal I ⊂ L, we can constructthe quotient algebra L/I. Seen as a
vectorspace,L/I issimplythequotientspace,i.e.
L/I ={x+I |x∈L},where
x+I ={x+i|i∈I}.
The bracket operation on L/I is then defined by [x+I,y+I] = [x,y]+I. It is easy to check that this is well
defined.
1.2 Nilpotent, solvable, simple and semisimple Lie algebras
Definition. ALiealgebraLiscalledsimple,if[L,L](cid:54)=0andLhasnoidealsexceptitselfand0.
Notethatthisimplies[L,L]=LandZ(L)=0.
TheLiealgebrasl (C)issimple:
2
Proof. Assume I isanon-zero idealofsl (R). Takex ∈ I, x (cid:54)= 0, thenx = αe+βf +γg, where {e, f, g} is
2
thebasisofsl (R)introducedabove,α, β, γ ∈Randatleastonecoefficientisnon-zero. Nownotethat
2
[[x,e],e]=−2βeand
[[x,f],f]=2αf.
Thus, if either α or β are non-zero, I = sl (R). On the other hand, if α = β = 0, then γ (cid:54)= 0 and, because
2
[x,e]=2γxand[x,f]=−2γy,I =sl (R).
2
Definition. Given a Lie algebra L, we define a sequence of ideals by L(0) = L, L(i+1) = [L(i),L(i)]. The Lie
algebraLiscalledsolvableifL(i) =0forsomei.
Asimplealgebraisnotsolvable(sinceL(i) =L∀i).
Definition. GivenaLiealgebraL,wedefineasequenceofidealsbyL0 =L,Li+1 =[L,Li]. TheLiealgebra
LiscallednilpotentifLi =0forsomei.
Note that, since L(i) ⊂ Li ∀i, all nilpotent Lie algebras are solvable. A simple algebra is never nilpotent
(becauseitisnotevensolvable). Abelianalgebrasarenilpotentandthereforealsosolvable.
Consider again the Heisenberg Lie algebra h . We have already seen that h1 = h(1) = [h ,h ] = (cid:104)z(cid:105) is
n n n n n
one-dimensional and therefore abelian, i.e. h is solvable with h(2) = 0. Since [x ,z] = [y ,z] = 0, it is also
n n i i
nilpotentwithh2 =0.
n
The non-abelian two-dimensional Lie algebra L with the basis {x, y} and [x,y] = x is an example for a
solvableLiealgebra,whichisnotnilpotent:
L(1) =L1 =[L,L]=(cid:104)x(cid:105),
5 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
L(2) =[L1,L1]=0,but
Ln =L1 ∀n≥1.
Definition. AnilpotentLiealgebraofdimensionniscalledfiliform,ifLn−2 (cid:54)=0(andthusLn−1 =0).
LetLbeavectorspacewithbasis{x , ..., x }. Ifwedefineabracketoperationby
1 n
[x ,x ]=x fori∈{2, ..., n−1},
1 i i+1
LbecomesafiliformLiealgebraandiscalledthestandardfiliformLiealgebraofdimensionn.
Definition. ALiealgebraLiscalledsemisimple,ifithasnonon-zerosolvableideals. Thisisequivalenttothe
condition,thatLisadirectsumofsimpleLiealgebras(hencethename).
Proposition2. IfLisaLiealgebraandL/Z(L)isnilpotent,thenLisalsonilpotent.
Proof. IfL/Z(L)isnilpotent,Ln ⊂Z(L)forsomen. ItfollowsthatLn+1 =0.
Proposition3. LetLbeaLiealgebra.
(i) IfS andS aretwosolvableidealsinL,thenS +S isasolvableidealinL.
1 2 1 2
(ii) IfN andN aretwonilpotentidealsinL,thenN +N isanilpotentidealinL.
1 2 1 2
Proof.
(i): Clearly the sum of two ideals is another ideal. Now note that (S +S )/S ∼= S /(S ∩S ) (isomorphism
1 2 2 1 1 2
theorem). TherightsideisahomomorphicimageofS andthereforesolvable,whichimpliesthattheleftside
1
is also solvable. It remains to show that, if I is a solvable ideal in a Lie algebra K and K/I is solvable, K is
solvableaswell. Assume(K/I)(n) =0,i.e. K(n) ⊂I. SinceI issolvable,I(m) =0forsomem. Itfollowsthat
K(n+m) =0.
Theprooffor(ii)isanalogous.
Definition. Let L be a Lie algebra. The preceeding proposition implies the existence of a maximal solvable
idealinLandamaximalnilpotentidealinL. ThemaximalsolvableidealiscalledtheradicalofLanddenoted
byRad(L). ThemaximalnilpotentidealinLiscalledthenilradicalanddenotedbyNil(L).
6 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
2 The representation theory of Lie algebras
Definition. ArepresentationofaLiealgebraLonavectorspaceV isaLiealgebrahomomorphismφ : L →
gl(V).
Definition. A representation is called faithful, if it is one-to-one, i.e. if the homomorphism φ : L → gl(V) is
injective.
2.1 Examples
Animportantexampleisthesocalledadjointrepresentation,definedby
ad:L→gl(L), x→ad ,where
x
ad :L→L, ad (y)=[x,y].
x x
IfLisaLiealgebraandx∈L,thenxiscalledad-nilpotentifad isnilpotent,i.e. (ad )n =0forsomen. The
x x
kernel of ad is the center of L. This means that, if Z(L) = 0, ad is injective and therefore isomorphic to its
image,whichisasubalgebraofgl(L). So,anysimpleLiealgebraisisomorphictoalinearLiealgebra.
Consideragainsl (R)withitsbasis{e, f, g}andthecommutationrelations
2
[e,f]=g, [e,g]=−2e, [f,g]=2f.
We are interested in the (faithful) representations of sl (R) on vector spaces V of different dimensions. For
2
dimV = 2, this is easy: a faithful representation is given by the identity map. Now consider V = R3 ∼= sl (R)
2
withbasis
     
1 0 0
0∼=e, 1∼=−g, 0∼=f
0 0 1
and the adjoint representation. Since sl (R) is simple, the adjoint representation is faithful, its matrix form is
2
givenby
     
0 2 0 0 0 0 2 0 0
ade =0 0 1, adf =1 0 0, adg =0 0 0 .
0 0 0 0 2 0 0 0 −2
As a last example consider the non-abelian two-dimensional Lie algebra with the basis {x, y} and [x,y] = x.
Notethateventhoughthisalgebraissolvable,itscenteriszeroandwecanagainusetheadjointrepresenta-
tion. Thematrixformis
(cid:18) (cid:19) (cid:18) (cid:19)
0 1 −1 0
ad = , ad = .
x 0 0 y 0 0
2.2 Modules, submodules, quotient modules
In the literature one often finds the notion of (left) L-modules, which is (in this context) basically just another
way of saying "representation of L". For completeness, even though in this thesis we will always speak of
representations,ashortsummaryofL-modulesispresentedinthefollowing.
Definition. Let L be a Lie algebra. A vector space V, together with an operation L×V → V, (l,v) → l.v, is
calleda(left)L-module,if
(i) (αl +βl ).v =α(l .v)+β(l .v),
1 2 1 2
(ii) l.(αv +βv )=α(l.v )+β(l.v ),
1 2 1 2
(iii) [l ,l ].v =l .l .v−l .l .v.
1 2 1 2 2 1
Givenarepresentationφ:L→gl(V),thevectorspaceV isan(left)L-modulewith
l.v =φ(l)(v).
7 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
Conversely,givenanL-moduleV,theaboveequationdefinesarepresentationφ:L→gl(V).
A L-submodule U is simply a subspace of V, which is closed w.r.t. the inherited operation, i.e. L.U ⊂ U.
Equivalentely, if φ : L → gl(V) is a representation of L and U is a subspace of L which is left invariant under
φ(L),i.e. φ(L)(U)⊂U,thenφ:L→gl(U)iscalledasubrepresentation.
V (cid:54)= 0 is called irreducible, if its only submodules are 0 and itself. V is called completely reducible, if it is
a direct sum of irreducible submodules. Again, irreducible/completely reducible representations are defined
equivalentely.
NotethatanyLiealgebraLisanL-modulefortheadjointrepresentationanditsL-submodulesareitsideals.
Therefore, any simple Lie algebra, seen as an L-module, is irreducible. A Lie algebra is called reductive, if
its adjoint representation is completely reducible, i.e. the Lie algebra is a direct sum of ideals. An equivalent
definitionisthataLiealgebraLisreductive,ifeveryidealI hasacomplementeryidealK,i.e. L=I⊕K.
Alinearmapψ :V →W,whereV andW areL-modules,iscalledahomomorphismofL-modules,if
ψ(l.v)=l.ψ(v).
2.3 Structure theorems: Lie and Engel
TheaimofthissubsectionistoprovethetheoremsofLieandEngel,butfirst,afewpreliminaryresultswillbe
derived.
Theorem1. Letx∈gl(V)benilpotent. Thenad isalsonilpotent.
x
Proof. Define
l :gl(V)→gl(V), l (y)=xy and
x x
r :gl(V)→gl(V), r (y)=yx.
x x
Ifxisnilpotent(e.g. xn =0),thensoarel andr :
x x
(l )n =xny =0,
x
(r )n =yxn =0.
x
Therefore,ad =l −r isnilpotentaswell(notethatl ◦r =r ◦l ):
x x x x x x x
(ad )2n =(cid:80)2n (cid:0)2n(cid:1)(l )2n−k(−r )k =0.
x k=0 k x x
Theorem2. LetV (cid:54)= 0beafinite-dimensionalvectorspaceandLasubalgebraofgl(V). IfallelementsofL
arenilpotent,thenthereexistsanonzerov ∈V,suchthatl(v)=0∀l∈L.
Proof. TheproofusesinductionondimL:
dimL=1: Denote the unique basis vector of L by l . Then choose an arbitrary vector v ∈ V and define the
B
sequence v = v, v = l (v ). Since l is nilpotent, v = 0 for some n. If we choose n minimal,
0 i+1 B i B n
v (cid:54)=0andl(v )=0∀l∈L.
n−1 n−1
dimL>1: Let K be a maximal proper subalgebra of L (such an algebra exists since L is finite-dimensional and
nilpotent). Let us now show that K is an ideal of L. Since all elements of L are nilpotent and ad ∈
k
gl(L)∀k ∈K,itfollowsfromtheorem1that
∀k ∈K ∃n:(ad )n =0,i.e. theimageofad:K →gl(L)isnilpotent.
k
ad alsoactsonthequotientspaceL/Kbyad (l+K)=ad (l)+K =[k,l]+K. Theinductionhypothesis
k k k
guaranteesustheexistenceofanvectorx∈L, x(cid:54)∈K,suchthat
ad (x+K)=0∀k ∈K,i.e. [x,k]∈K ∀k ∈K.
k
This means K is properly included in N (K) and since K is maximal, N (K) = L, i.e. K is an ideal in
L L
L.
8 HoferJoachim
AnintroductiontoLiealgebrasandthetheoremofAdo
K+(cid:104)x(cid:105),where(cid:104)x(cid:105)denotesthespanofx,isasubalgebraofLand,sinceK isalreadyamaximalproper
subalgebraandx(cid:54)∈K,itfollowsthat
L=K+(cid:104)x(cid:105).
Duetotheinductionhypothesisthereisav ∈V,s.t. K(v)=0,i.e. k(v)=0∀k ∈K. Define
W ={v ∈V|K(v)=0}=(cid:54) 0.
Itnowsufficestoshowtheexistenceofav ∈W −{0}withx(v)=0. NotethatW isstableunderL,i.e.
(cid:0) (cid:1) (cid:0) (cid:1)
L(W)={l(w)|l∈L, w ∈W}⊂W,sincek l(w) =l k(w) −[l,k](w)=0.
In particular, W is stable under (cid:104)x(cid:105) and therefore (cid:104)x(cid:105)| ⊂ gl(W). It follows (again by the induction
W
hypothesis)that
∃v ∈W : (cid:104)x(cid:105)(v)=0andthereforeL(v)=0.
WiththeseresultswearenowpreparedtoproveEngel’stheorem.
Engel’stheorem. IfallelementsofaLiealgebraLaread-nilpotent,thenLisnilpotent.
Proof. Again, by induction on dimL, the case L = 0 being trivial. The image of ad : L → gl(L) consists of
nilpotentendomorphismsandthereforesatisfiestherequirementsoftheorem2. ItfollowsthatZ(L)(cid:54)=0. Now
(cid:0) (cid:1)
L/Z(L) consists of ad-nilpotent elements and dim L/Z(L) < dimL. By the induction hypothesis, L/Z(L) is
nilpotent. Duetoproposition2Lisalsonilpotent.
Proposition4. Undertheassumptionsoftheorem2,thereexistsabasisofV,relativetowhichthematrices
ofLarestrictlyuppertriangular.
Proof. Let v ∈ V be a non-zero vector, such that L(v ) = 0 (the existence of such a vector is guaranteed by
0 0
theorem 2. Define W = V/(cid:104)x (cid:105). Note that L| still consists of nilpotent endomorphisms. Using induction on
0 W
dimV (thecasedimV =0beingtrivial),W hasabasis{v , ..., v }(modv )relativetowhichL| consists
1 n−1 0 W
ofstrictlyuppertriangularmatricesand{v , ..., v }isthebasisofV wewerelookingfor.
0 n−1
Theorem 3. Let V (cid:54)= 0 be a finite-dimensional vector space over the field F and L a solvable subalgebra of
gl(V). ThenV containsacommoneigenvectorforallelementsofL.
Proof. TheproofisdonebyinductionondimLandissimilartotheproofoftheorem2.
dimL=1: There is one basis vector l with an eigenvalue λ (F is algebraically closed). The corresponding eigen-
B
vectorisaneigenvectorforallelementsα·l inLwitheigenvaluesα·λ.
B
dimL>1: SinceLissolvable,[L,L]isaproperidealinL.
Therefore,asubspaceK ofcodimensiononecontaining[L,L]exists,i.e. K isanidealand
∃x∈L−K : L=K+(cid:104)x(cid:105).
The induction hypothesis guarantees the existence of a common eigenvector for all elements in K, i.e.
thereexistv ∈W andalinearfunctionλ:K →F,suchthatk(v)=λ(k)·v foreveryk ∈K. Therefore
W ={v ∈V|∃λ:K →F: k(v)=λ(k)·v ∀k ∈K}=(cid:54) 0.
Assume for a moment that L(W) ⊂ W. It follows, that (cid:104)x(cid:105)| ⊂ gl(W). Since the field is algebraically
W
closed,xhasaneigenvectorv ∈W,whichisthusalsoaneigenvectorforeveryl∈L=K+(cid:104)x(cid:105).
ItremainstoshowthatLleavesW invariant. Notethatforarbitraryk ∈K,l∈L,w ∈W
(cid:0) (cid:1)
k l(w) =(l◦k)(w)−[l,k](w)
andthus,sinceK isanidealinLand[l,k]∈K,
(cid:0) (cid:1)
k l(w) =λ(k)·l(w)−λ([l,k])·w.
It therefore suffices to show that λ([l,k]) = 0 for all k ∈ K, l ∈ L. Fix w ∈ W, l ∈ L and choose n ∈ Z+
minimal,suchthatw, l(w), ..., ln(w)arelinearlydependent. Define
W =0, W =(cid:104)w, ..., li−1(w)(cid:105).
0 i
9 HoferJoachim